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Students can Download Chapter 6 Non-Competitive Markets Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Non-Competitive Markets

Plus Two Economics Non-Competitive Markets One Mark Questions and Answers

Question 1.
Point out the value of the Marginal Revenue when the demand curve is elastic.
Answer:
When the demand curve is elastic, the value of Marginal Revenue (MR) is zero.

Question 2.
Identify from the following demand curve faced by a firm under monopolistic competition.
Answer:

Answer:
(C)

Question 3.
The form of the market having only 2 sellers is called:
(a) monopoly
(b) duopoly
(c) oligopoly
(d) monopolistic competition
Answer:
(b) duopoly

Question 4.
Differentiated products is a characteristic of
(a) Monopolistic competition only
(b) Oligopoly only
(c) Both Monopolistic competition & Oligopoly
(d) Monopoly
Answer:
(c) Both Monopolistic competition & Oligopoly

Plus Two Economics Non-Competitive Markets Two Mark Questions and Answers

Question 1.
Suppose there are only two firms manufacturing cars in India, namely, Maruti and Hyundai. What market form is this?
Answer:
lip This market form is known as duopoly market. Duopoly is a market structure in which there are only two firms producing a product.

Question 2.
What do you mean by cartel?
Answer:
Cartel is a kind of mutual agreement or coordination of the output and pricing policies of firms having then individual identities so as to act if it is a monopoly. This is a feature of oligopoly market.

Question 3.
Observe the following figures and identify the market situations.

Answer:

1. Monopoly
2. Monopolistic competition

Question 4.
Suggest any 2 examples of a monopolistically competitive market.
Answer:

1. Soap industry
2. Toothpaste industry.

Question 5.
The features of firms under different market structure is given below. Classify them into perfect competition and oligopoly.
(a) Collusion
(b) Free entry and exit
(c) Intertdependance
(d) Firms are price takers
Answer:
a & c Oligopoly, b & d Perfect competition

Plus Two Economics Non-Competitive Markets Three Mark Questions and Answers

Question 1.
The monopolist cannot determine the price and quantity simultaneously. Do you agree? Substantiate your answer.
Answer:
Yes, I agree to the statement that the monopolist cannot determine the price and quantity simultaneously. This is because, if the monopolist wants to sell more of the commodity, he needs to reduce the price. Therefore, he can change either of the price or the quantity.

Question 2.
Examine the relationship between Marginal Revenue and Price Elasticity of Demand.
Answer:
Whenever the MR is positive the price elasticity of AR (Demand Curve) isgreaterthan one, that is elastic. When the MR is zero the price elasticity of AR (Demand Curve) is 1, that is unitary elastic. When the MR is negative the price elasticity of demand curve is less than one, that inelastic.

Question 3.
Choose the correct answer
a. In monopoly:

1. There are many producers
2. There is no seller
3. There is no buyer
4. There is single seller

b. When two commodities are complementary to one another

1. They may be jointly demanded
2. They may be complementary goods
3. They may be substitutes
4. None of the above

c. Generally government fix control price

1. Equal to equilibrium price
2. Less than equilibrium price
3. More than equilibrium price
4. None of these

Answer:
a. 4. There is single seller
b. 1. They may be jointly demanded
c. 2. Less than equilibrium price

Question 4.
State whether the followirig satements are true or false.

1. The seller in a monopoly is a price maker.
2. Price leadership is an important feature of oligopoly.
3. Selling cost is the cost of producing the commodity.

Answer:

1. Time
2. True
3. False. Selling cost is the cost of selling a Product.

Question 5.
State the condition and long run equilibrium in a monopoly competitive industry.
Answer:
The long run equilibrium conditions in a monopolisti-cally competitive industry are:
MR = LMC
P = LAC but P > LMC MR = LMC
P = LAC, P > LMC

Question 6.
The diagram below shows the equilibrium condition of a zero cost monopolist. Find out the quantity produced by such a firm, explain

Answer:
The firm will produce oq1 level of output to maximise the profit. Because at this level of output, the firm satisfies a condition that is MR = MC. Since the firm faces zero cost its MR also will be zero. For this equilibrium, MR and MC should be equal. Since MC is zero in all level of output the firm will produce at a level where its MR is zero.

Question 7.
Examine the behaviour of average revenue and marginal revenue of a firm which can sell more units of a good only by lowering the price of that good. Explain with the help of a diagram.
Answer:
monopolist can sell more units of the good only by lowering the price. Therefore AR & MR will be downward sloping. Draws the appropriate diagram.

Question 8.
Categorize the following features under two headings Perfect Competition and Monopolistic Competition. 4 Large Number of Producers, Differentiated products, Some Pricing power, Productive Efficiency in the Long run, Low Barriers, Homogeneous products, Long run Price = MC, Many producers, Zero Barriers, Productive Inefficiency in the Long run, Price Takers, Long run Price >MC.
Answer:
1. Perfect competition

• Large Number of Producers.
• Productive Efficiency in the Long run
• Homogeneous products
• Long run Price = MC
• Zero Barriers
• Price Takers

2. Monopolistic competition

• Differentiated products
• Some Pricing power
• Low Barriers,
• Many producers
• Productive Inefficiency in the Long run
• Long run Price >MC.

Question 9.
Consider the commodities given below. Identify the most likely market situation in which they are produced. Substantiate.

1. Airline industry.
2. Potatoes.
3. Toilet soap.

Answer:

1. Oligopoly – only a few producers
2. Perfecly Competitive Market – large number of producers
3. Monopolistic Competition – Many sellers producing differentiated products

Question 10.
A table related to a particular market is given below:

1. Find AR & MR
2. Identify the market related to the table.
3. Establish the relationship between TR, AR & MR

Answer:
1.

2. Monopoly.

3. relationship between TR, AR & MR

• TR rises and then falls
• AR is always falling but lies above MR
• MR falls and becomes negative

Question 11.
The total revenue equation of a firm is given by the equation,TR = 20Q – 2Q²

1. Calculate TR, AR & MR.
2. Identify the market form related to this equation

Answer:
1. TR = 20Q – 2Q2
\(\begin{aligned}
&A R=\frac{T R}{Q}=\frac{20 Q-2 Q^{2}}{Q}=20-2 Q\\
&\mathrm{MR}=\frac{\delta \mathrm{TR}}{8 \mathrm{Q}}=\frac{\left(20 \mathrm{Q}-2 \mathrm{Q}^{2}\right)}{\mathrm{Q}}=20-4 \mathrm{Q}
\end{aligned}\)

2. Monopoly

Plus Two Economics Non-Competitive Markets Five Mark Questions and Answers

Question 1.
State whether the following statements are true or false. Rewrite the statements if they are wrong.

1. The products in perfect competition are heterogeneous
2. The seller in monopoly is a price maker
3. Price leadership is an important feature of oligopoly.
4. Duopoly is a market situation in which two buyers buy the commodity
5. Selling cost is the cost for producing the commodity.

Answer:

1. False. Products in perfect competition are homogenous
2. True
3. True
4. False. Duopoly is a market situation in which two sellers supply the commodity
5. False. Selling cost is the cost for selling or giving publicity for the commodity

Question 2.
Find odd one out.

1. Single seller, price maker, selling cost, control over supply
2. Fairly large number of firms, product differentiation, selling cost, price maker
3. A few firms, interdependence between firms, no transport cost, indeterminate demand curve
4. Tata steel, Reliance industries, Post and Telegraph

Answer:

1. Selling cost. Others are features of monopoly
2. Price maker. Others are features of monopolistic competition
3. No transport cost. Others are features of oligopoly
4. Post and Telegraph. Others are private sector companies.

Question 3.
Match column B and Q with column A.

Answer:

Question 4.
The market price, quantity and total cost of a firm are given in the following table. Find out.

1. MRandMC
2. Equilibrium price and equilibrium quantity
3. TR, TC and Total profit at equilibrium

Answer:
1.

2. Equilibrium price is Rs. 19 and quantity is 6.

3. At equilibrium
TR= 114
TC = 109
Total profit = 5

Question 5.
Categorize the following into different market forms.

1. Indian Railways
2. Toothpaste
3. Hero Honda
4. KSEB

Answer:

1. Indian Railways – monopoly
2. Toothpaste – monopolistic competition
3. Hero Honda – oligopoly
4. KSEB – monopoly

Question 6.
Prepare a note on price rigidity.
Answer:
Price rigidity is an important feature of oligopoly. Price rigidity means that price will remain rigid without much fluctuation. This is because, price increase by one firm will not be followed by other firms.

However, price reduction by one firm will be followed by other firms, due to this; the firm affecting price change will not get the benefits from the reduction of price. Therefore, no firm will reduce or increase the price. This leads to a situation of price rigidity in the oligopoly markets.

Question 7.
Assume that there are two firms A and B in a duopoly market. Firm B supplies zero output. Firm A realizes that maximum demand in the market is 20 units, and he supplies half of it, i.e., 10 units. Construct a table the different steps showing the quantity supplied by the firms.
Answer:

Question 8.
Classify the statements under the head features of oligopoly market.
Answer:

1. Large no. of buyers and sellers
2. Firm is a price maker
3. Few number of sellers
4. Products may be homogneous or differentiated.
5. There is no interdependence between firms
6. There are no barries to entry
7. Firm is price taker
8. Interdependence between firms
9. Entry restricted
10. Price rigidity prevails oil

Answer:
The features of oligopoly market are given below:

1. Few number of sellers
2. Product may be homogneous or differentiated
3. There are no barriers to entry
4. Interdependence between firms
5. Price rigidity prevails.

Question 9.
Price rigidity is an important feature of oligopoly. Can you explain what is price rigidity.
Answer:
Price rigidity is an important feature of oligopoly market. In oligopoly market, price does not change m easily in response to change in demand. If one firm decides to increase the price to earn high profit and the other firms do not do so, due to increase the price, the demand of product will fall and it causes fall in revenue and profit. Hence it is not rational for any firm of increase the price. Thus in an oligopoly market, price remain rigid.

Question 10.
Suppose that firm A enters in a duopoly market for production of commodity X at zero cost. He finds that the total demand forX in the market is 300 units. When he starts production, firm B enters in market. Find out the profit maximising quantity by each firm.
Answer:
In order to maximisejprofit each firm will produce 1/3 of the total markerdemand. In one example, total demand in the market is 300 units. Therefore, the profit maximising output is 1/3 x 300 = 100 units.

Question 11.
Identify the market structure.

Description	Concept
Market with a few firm
Market with only two firms
Market with only one buyer
Market with only one seller

Answer:

Description	Concept
Market with a few firm	Oligopoly
Market with only two firms	Duopoly
Market with only one buyer	Monopsony
Market with only one seller	Monopoly

Question 12.
Make pairs.
Price maker, Price leadership, Monopoly, Monopsonist, single buyer, Oligopoly, monopolistic competition, selling cost.
Answer:

• Price maker – Monopoly
• Price leadership – Oligopoly
• Single buyer – Monopsonist
• Selling cost – Monopolistic competition.

Question 13.
The demand curves of different market situations are given below.

1. Identify market situations represented by each demand curve.
2. Give reasons for the different shapes of demand curves in these two market forms.

Answer:
1. Figure (1) represents perfect competition market Figure (2) represents monopoly market.

2. In perfect competition, there are large number of buyers and sellers. Each firm is a price taker and there is uniform price prevailing in the market. Since each unit is sold at uniform price, P = MR = AR in the market. Therefore, demand curve is horizontal straight line. However, in a monopoly market, firm is a price maker. He can vary the price. If he wants to sell more of the product, he need to reduce the price. Therefore, the demand curve is falling downward.

Question 14.
Prepare a note on monopolistic competition.
Answer:
A market structure where the number of firms is large, there is free entry and exit of firms, but the goods produced by them are not homogeneous such a market structure is called monopolistic competition. Its features are as following:

1. There are large number of buyers and sellers.
2. There is free entry and exist in long run.
3. There is product differentiation.

The monopolist produces less and charges a higher price compared to perfect competiton. It is found in the industry where there is large number of sellers, selling differentiated but close substitute products. Monopolistic competition in a commodity market arises due to the commodity being non-homogeneous.

Question 15.
Compare the price and output of a firm under perfect competition and monopolistic competition.
Answer:
A firm under perfect competition is a price taker and have a horizontal demand, but a firm under monopolistic competition is a price maker and faces a demand curve that in downward sloping and elastic. Under perfect competition MR = AR. So the firm produces more output and charge less compared to monopolistic competition. Under monopolistic MR < AR.

Question 16.
Identify the market condition with following feature.

1. Interdependence
2. Price rigidity
3. Entry restriction

Explain why prices are rigid in such market situations.
Answer:
Oligopoly. Under such markets, the price is supposed to be rigid. The reason is that here the firms are interdependent. The actions of every firm will be determined by the actions and reactions of every other firm. If one firm increases the price none other follows. The customers of that firm may switch to other firms. The firm which increased the price may feel a fall in revenue and profit.

On the other hand, if one firm reduces the price everyone else will follow. All firm’s revenue and profit fall. So firms under oligopoly will always try to keep their price rigid.

Question 17.
Two tables are given below.

1. Find TR, AR&MR.
2. There are five variables in each of the above table. Which two variables in both the tables have the same values? Give reasons.
3. The two tables are related to two market forms. Identify the form of market for each table. Give reasons.

Answer:
1.

2. Table 1: Price and AR are same Table 2: Price, AR, and MR are same

3. Table 1: Monopoly market because by reducing price firm sells more
Table 2: Perfect competition. Firm is price takes and P=AR=MR

Question 18.
Does the statement below better describe a firm operating in a Perfectly Competitive market or a firm that is Monopoly?

1. The demand curve faced by the firm is downward sloping.
2. The demand curve and the MR curve are the same.
3. Entry and exit are relatively difficult.
4. Price Taker
5. Price Maker

Answer:

1. Monopoly
2. Perfectly Competitive Market
3. Monopoly
4. Perfectly Competitive Market
5. Monopoly

Plus Two Economics Non-Competitive Markets Eight Mark Questions and Answers

Question 1.
Prepare a seminar paper on ‘Non-Competitive Markets’.
Answer:
Respected teachers and dear friends,
The topic of my seminar paper is noncompetitive markets. As we know there are different kinds of markets depending upon the number of firms, nature of the product, freedom of entry and exit, etc. On the basis, of the above, we name the non competitive markets as monopoly, monopolistic competition and oligopoly. .

Introduction:
A market structure in which there is a singe seller is called monopoly. A market structure where the . number of firms is large, there is free entry and exit of firms, but the goods produced by them are not homogeneous. Such a market structure is called monopolistic competition. If the market of a particular commodity consists of more than one seller but the number of sellers is few, the market structure is termed oligopoly.

Contents:

1. Monopoly market
2. Monopolistic competition
3. Oligopoly

1. MONOPOLY MARKET:
Monopoly may be defined as a market situation in which there is only a single seller. He controls the entire market. The term monopoly has derived from two Greek words such as ‘mono’ means single and poly means ‘seller’. The meaning of the combined term is single seller. In a boardersense, a monopolist is single seller of a commodity which does not have close substitutes, e.g. KSEB

Features of Monopoly Market:
Some of the salient features of monopoly are as follows:

1. There is only a single firm producing the product
2. There is no close substitute for the product
3. Entry is denied for other producers
4. Since there is only one seller, the firm and the industry are same
5. The firm under monopoly is the price maker

2. MONOPOLISTIC COMPETITION:
Monopolistic competition is a market characterized by the elements of perfect competition and monopoly. It is a market situation characterized by large number of firms producing various kinds of goods and services. The products of a firm will be different from the products of other firms in terms of size, shape, smell, colour, etc.

Features
The salient features of perfect competition are as follows:
1. Large number of buyers and sellers:
Under monopolistic competition, there exists large number of buyers and sellers. But the number of sellers will be less compared to perfect competition.

2. Product differentiation:
One of the most important characteristic of monopolistic competition is the existence of product differentiation. Each firm has its own product with unique brand names. The products of one firm will be different from the products of other firms in terms of size, shape, smell, color, etc.

3. Freedom of entry and exit:
Under monopolistic competition, there is freedom of entry and exit.

4. Selling cost:
The cost incurred for sales promotion such as advertisement, coupons, gifts, etc. are known as selling cost. Under monopolistic competition, the selling costs would be relatively high.

3. OLIGOPOLY:
The term oligopoly has derived from two terms oligo (small) and poly (seller). Thus oligopoly is a market situation characterized by competition among few sellers. In simple terms, it is a competition among few sellers in the market selling either homogenous or differentiated product. The industries manufacturing car, motorcycle, scooter, etc. are some of the examples for oligopolistic competition.

The main features of oligopolistic competition are as follows:
1. Few sellers:
The number of sellers or producers would be few under oligopolistic competition.

2. Homogneous or differentiated products:
The products sold under oligopolistic competition would be either homogneous (e.g. gas, petrol) or differentiated (e.g. car, scooter)

3. Free entry and exit:
Free entry and exit persist under oligopolistic competition.

4. Selling cost:
Firms spend on advertisement and sales promotion.

5. Interdependence of the firms:
Since the number of firms under oligopoly are few, they are highly interdependent. The action of one firm will certainly have impact on other firms in terms of price, quality of the product, etc.

6. Price leadership:
Some of the firms may emerge as price leaders under oligopoly. The price leader could be the first firm in the industry or the firm with largest number of consumers. The price leader takes important decisions regarding vital decisions such as the price of the product or number of units to be produced in the market, etc.

Conclusion:
Thus it can be concluded that there are three kinds of non-competitive markets. This classification is made on the basis of the number of firms, nature of the product, freedom of entry and exit, etc. In contrast to perfect competition, we find that these market forms are more realistic.

Question 2.
Prepare a table to show the distinction between monopoly, monopolistic competition, and oligopoly. Major points of distinction are given below in the table.

Answer:

Question 3.
If the market of a particular commodity consists of more than one seller but the number of sellers is few, the market structure is termed oligopoly. The special case of oligopoly where there are exactly two sellers is termed duopoly. We shall explain the different ways in which the oligopoly firms may behave.

1. Firstly duopoly firms may collude together and decide not to compete with each other and maximize total profits of the two firms together. In such a case the two firms would behave like a single monopoly firm that has two different factories producing the commodity.

2. Secondly, take the case. of a duopoly where each of the two firms decide how much quantity to produce by maximizing its own profit assuming that the other firm would not change the quantity that it is supplying. We can examine the impact using a simple example where both the firms have zero cost.

3. Thirdly, some economists argue that oligopoly market structure makes the market price of the commodity rigid, i.e., the market price does not move freely in response to changes in demand.

Question 4.
Name important non-competitive markets and give the meaning of them.
Answer:
The important forms of non-competitive markets also:

1. Monopoly
2. Monopolistic competition
3. Oligopoly

1. Monopoly:
A monopoly is a market situation in which there is a single seller of the commodity and no close substitutes of the commodity are available. The single seller can influence the price by varying his sales.

2. Monopolistic Competition:
Monopolistic competition is a market situation in which both the monopolistic element and the competitive elements are present. Its basic features are large number of buyers and sellers in the market and existence of differentiated products.

3. Oligopoly:
Oligopoly is market situation in between monopolistic competition and monopoly. In this market form, there are only a few sellers of the commodity and each seller has a substantial share in the market.

Question 5.
The diagram below shows the level of output produced and price charged in monopoly and perfect competition.

1. Identify the levels of output and price charged in monopoly and perfect competition, explain.
2. Critically evaluate the merits and demerits of perfect competition.

Answer:
1. A firm, if it is under monopoly, will produce when its MR = MC and will charge a price which is equal to AR. It produces oq level of output and charges a price op. But if it is in perfect competition if the market would

2. Confirmed to monopoly perfect competition would charge higher prices and produce less quantity. It is argued that the monopoly firms benefit themselves at the cost of consumers. The monopolist may get a profit even in tire long run and the consumers pay more and get less quantity.

But it is another argument. The profit made by the monopolist would be used for research and development and it may be useful for society in the long run in terms of new technology and new products. Moreover, due to the economies of scale the cost of the monopolist may be much lower than the cost of a firm under perfect competition.

Students can Download Chapter 9 Coordination Compounds Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 9 Coordination Compounds

Plus Two Chemistry Coordination Compounds One Mark Questions and Answers

Question 1.
The ions or molecules bound to the central atom/ion in the coordination entity are called ___________.
Answer:
Ligands

Question 2.
In [Co(C₂O₄)₃]^3-, the coordination number of cobalt is _________.
Answer:
six

Question 3.
Which complex has a square planar structure?
(a) [Ni(CO)₄]
(b) [NiCI₄]^2-
(c) [Ni(H₂O)₆]²⁺
(d) [CU(NH₃)₄]²⁺
Answer:
(d) [CU(NH₃)₄]²⁺

Question 4.
Say TRUE or FALSE.
[Co(NO₃)(NH₃)₅]SO₄ and [Co(NO₃)(NH₃)₄(SO₄)](NH₃) are ionisation isomers.
Answer:
FALSE

Question 5.
The formula of Wilkinson’s catalyst is _________.
Answer:
[RhCl(PPh₃)₃]

Question 6.
The charge of Ni in [Ni(CO)₄] is
(a) +1
(b) +2
(c) 0
(d) +4
Answer:
(c) 0

Question 7.
The central metal ion present in chlorophyll?
(a) Fe²⁺
(b) Cu²⁺
(c) Mg²⁺
(d) CO²⁺
Answer:
(c) Mg²⁺

Question 8.
EDTA is a dentate ligand
(a) uni dentate
(b) bidentate
(c) Tridentate
(d) hexadentate
Answer:
(d) hexadentate

Question 9.
Which is an example for homoleptic complexes
(a) [Co(NH₃Cl₂]⁺
(b) [CoNH₃)₆]³⁺
(c) [Cr(NH₃)(H₂O)₃]Cl₃
(d) [CoCl₂(en)₂]
Answer:
(b) [CoNH₃)₆]³⁺

Question 10.
Ammonia will not form complex with
(a) Ag²⁺
(b) Pb²⁺
(c) Cu²⁺
(d) Cd²⁺
(e) Fe²⁺
Answer:
(b) Pb²⁺

Plus Two Chemistry Coordination Compounds Two Mark Questions and Answers

Question 1.
In a seminar, Jishnu argued that the “hexaflourocobaltate(III) ion is highly paramagnetic than hexacyanoferrate(III) ion.

1. Do you agree with this words?
2. Explain it.
3. Write the formulae of the given coordination compounds.

Answer:

1. Yes
2. CN^– is a strong field ligand so paring occurs. Number of unpaired electron decreases, paramagnetism decreases. F^– is a weak field ligand so no paring occurs, paramagnetism increases.
3. [CoF₆]^3-and [Co(CN)₆]^3-

Question 2.
Raju: Coordination compounds are coloured.
Ramu: No, co-ordination compounds are colourless.

1. Whose statement is correct?
2. Explain the reason for your answer.

Answer:

1. Raju’s statement is correct. Coordination compounds are usually coloured.
2. The colour of coordination compounds is due to d-d transition.

Question 3.
Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands coordinated with the central metal ion?
Answer:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other.

Question 4.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu²⁺ ion. Explain why?
Answer:
FeSO₄ does not form any complex with (NH₄)₂SO₄. Instead, it forms a double salt FeSO₄. (NH₄)₂SO₄.6H₂O which dissociates completely into ions. CuSO₄ when mixed with NH₃ forms a complex [CU(NH₃)₄]SO₄ in which the complex ion [CU(NH₃)₄]²⁺ does not dissociate to give Cu²⁺ ion.

Question 5.
Write the geometrical isomers of [Pt(NH₃)(Br)(Cl) (py)] and how many of these will exhibit optical isomerism?
Answer:
The complex [Pt(NH3)(Br)(Cl)(py)] will form three geometrical isomers:

Square planar complexes of this type will not show geometrical isomerism.

Question 6.
Aqueous copper sulphate solution (blue in colour) gives:

1. a green precipitate with aqueous potassium fluoride and
2. a bright green solution with aqueous potassium chloride. Explain these experimental results.

Answer:
Aqueous copper sulphate contains a coordination entity, [Cu(H₂O)₄]²⁺ which is blue in colour. Water molecule is a weaker ligand than Cl^– and F^–.
1. On addition of aqueous KF solution, a new complex entity is formed wich is green in colour.

2. On addition of aqueous solution of KCl, an another bright green complex entity is formed which is soluble in water.

Question 7.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S(g) is passed through this solution?
Answer:
Aqueous solution of copper sulphate contains Cu²⁺ ions in the form of complex entity, [Cu(H₂O)₄]²⁺ and H₂O ligand is a weak ligand. When excess of KCN is added, a new coordination entity, [Cu(CN)₄]^2- is formed due to the following reaction:

Cyanide ligand, CN^– is a strong field lignad and stability constant of [Cu(CN)₄]^2- is quite large and thus practically no Cu²⁺ ions are left in solution. On passing H₂S gas, no CuS is formed due to non- availability of Cu²⁺ ions in solution.

Question 8.
Optical isomerism is usually exhibited by complexes containing polydentate ligand. What do you mean by ligand?
Answer:
Ligand is a neutral molecule or charged ion which can donate a lone pair of electron to the metal.

Question 9.
Coordination complexes are of different types. Name the compounds.

1. [Cr(H₂O)₅Cl₂]
2. K₃[Cr(C₂O₄)₃]

Answer:

1. Pentaaquadichloridochromium(II)
2. Potassiumtrioxalatochromate(III)

Question 10.
Write the IUPAC names of the following compounds.

1. K_3M[Fe(CN)₆]
2. [C0(NH₃)₅(CO₃)]Cl

Answer:

1. Potassiumhexacyanoferrate(III)
2. Pentaamminecarbanatocobalt(III) chloride

Question 11.
[Fe(CN)₆]^3- is paramagnetic, while [Fe(CN)₆]^4- is diamagnetic. Explain with the help of VB theory.
Answer:
In [Fe(CN)₆]^3- iron is in +3 state and in [Fe(CN)₆]⁴, iron is in +2 state. [Fe(CN)₆]^3- contains five electrons in d-level (3d⁵). In this complex iron undergoes d²sp³ hybridisation.

Due to the presence of one unpaired electron, [Fe(CN)₆]^3-, is paramagnetic. In [Fe(CN)₆]^4- iron contains six electrons in d-level (3d⁶). It undergoes d²sp³ hybridisation and has no unpaired electrons. Hence, [Fe(CN)₆]^4- is diamagnetic.

Plus Two Chemistry Coordination Compounds Three Mark Questions and Answers

Question 1.
Look at the following two diagrams.

1. Is the diagrams I and II correct? Justify. If the figure is not correct, redraw it.
2. Which theory is related to this?
3. Explain briefly, how this theory is applicable to octahedral complexes.

Answer:
1. No, Figure (II) is wrong.

2. Crystal field theory.

3. In the case of octahedral complexes, the ligands are approaching the ‘d’ orbitals through the axis. As a result of this the energy of d_x²-y² and d_z² orbitals increases and the energy of the remaining three orbitals decreases. The orbitals which possess high energy are represented as ‘e_g’ levels and the orbitals which possess less energy are represented as “t_2g” levels.

Question 2.
A list of coordination compounds are given below:
[Cr(H₂O)₆] Cl₃, [Co(NH₃)₅Br] SO₄, [Co(NH₃)₅NO₂]²⁺ and [Pt(NH₃)₂Cl₂]. Which type of isomerism do these compounds exhibit?
Answer:
Hydrate Isomerism, Ionisation Isomerism, Linkage Isomerism, Geometrical Isomerism.

Question 3.
The following are examples of coordination compounds. Identify the type of isomerism exhibited by each of them and write their possible isomers,

1. [Cr(NH₃)₅Br]SO₄
2. CrCl₃.6H₂O
3. [PtCl₂(NH₃)₂]

Answer:

1. Cr(NH₃)₅Br]SO₄ – Ionisation isomerism – [Cr(NH₃)₅SO₄]Br
2. CrCl₃.6H₂O – Hydrate isomerism – [Cr(H₂O)₅Cl]Cl₂.H₂₎
3. PtCl₂(NH₃)₂] – Geometrical isomerism

Question 4.
What will be the correct order for the wavelengths of absorption in the visible region for the following:
[Ni(NO₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺
Answer:
The order of the ligand in the spectrochemical series is H₂O < NH₃ < NO₂^–. Hence the wavelength of the light observed will be in the order:
[Ni(H₂O)₆]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(NO₂)₆]^4-

Thus, the wavelength of light absorbed (E = \(\frac{\mathrm{hc}}{\lambda}\)) will be in the opposite order:
[Ni(H₂O)₆]²⁺ > [Ni(NH₃)₆]²⁺ > [Ni(NO₂)₆]^4-

Question 5.
In a classroom discussion, Sajan argued that CN^–, OH^–, Cl^– etc. are examples for neutral ligands.

1. Do you agree with his argument?
2. If not, give a reason with the help of examples.
3. What do you mean by chelating ligand and chelation?

Answer:

1. No.
2. They are charged ligands.
3. If a polydentate ligand is coordinated to the metals, a ring structure is obtained. It is called chelate and the phenomenon is called chelation.

Question 6.
Ligands can be arranged according to the magnitude of Δ₀ and the arrangement is given below:
l- < Br^– < Cl^– < F^– < OH^– < H₂O < NH₃ < (en)

1. What is this series known as?
2. Is the sequence incorrect order?
3. Identify the weak field and strong field ligands.

Answer:

1. Spectrochemical series
2. Yes
3. The ligands above water are strong ligand and the ligands below water are weak ligands.

Weak filed ligands – l^–, Br^–, Cl^–, F^–, OH^–, H₂0 Strong filed ligands – NH₃, en

Question 7.
Consider the following coordination compounds.

• [Co(NH₃)₅ Cl] SO₄
• [Co(NH₃)₅ SO₄]Cl

1. Write down the IUPAC name of these compounds.
2. Name the isomerism exhibited by these compounds.

Answer:
1. The IUPAC name of compounds

• Pentaamminechloridocobalt(III) sulphate
• Pentaamminesulphatocobalt(III) chloride

2. Ionisation isomerism

Question 8.
Consider the following compounds:
[Co (NH₃)₅ NO₂] Cl₂ and [Co (NH₃)₅ ONO] Cl₂

1. Identify the isomerism exhibited by these compounds.
2. Explain ionisation isomerism with example.

Answer:

1. Linkage isomerism
2. This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion.
   e.g. [Co(NH₃)₅Cl] SO₄ & [Co(NH₃)₅SO₄] Cl

These complexes ionises as:

• [Co(NH₃)₅Cl] SO₄ → [Co(NH₃)₅Cl]²⁺ + SO₄^2-
• [Co(NH₃)₅SO₄] Cl → [Co(NH₃)₅SO₄]+ + Cl^–

Question 9.
Consider the statement: Crystal Field Theory (CFT) is applicable to octahedral and tetrahedral complexes.

1. Is this statement true?
2. Explain the crystal field splitting in octahedral complexes with the help of a neat diagram.

Answer:

1. Yes
2. In an octahedral crystal field the ligands are approaching the metal along the axes. Hence, the energy of d_x²-y² and d_z² orbitals (e_g set) increases by 3/5Δ₀and that of d_xy, d_yz and d_xz orbitals (t_2g set) decreases by 2/5 Δ₀

Question 10.
When a ligand approaches to an octahedral complex, the degenerate ‘d’ orbitals undergoes splitting.

1. What will be the observation?
2. What are the factors influencing crystal field splitting energy?

Answer:
1. The energy of d_x²-y² and d_z² orbitals (e_g set) increases by 3/5Δ₀ and that of d_xy, d_yz and d_xz orbitals (t_2g set) decreases by 2/5 Δ₀

2. The factors influencing crystal field splitting energy.

• Nature of the ligand
• Geometry of the complex
• Valency of the metal

Question 11.
What will happen when a ligand approaches to a tetrahedral complex?
Answer:
the energy of d_xy, d_yz and d_xz orbitals (t_2g set) increases by 2/5Δ_t and that of d_x²-y² and d_z² orbitals (e_g set) decreases by 3/5 Δ_t.

Question 12.
Consider the complex ion [Ti (H₂O)₆]³⁺ In the case of an octahedral complex, what is the condition for the pairing of forth electron in the d- level?
Answer:
If the crystal field splitting energy is greater than the pairing energy, the fourth electron will pair at the t_2g level and if the pairing energy is greater than the crystal field splitting energy the electron will go to the on e_g level.

Question 13.
Is bidentate ligands same as the amidentate ligands? Justify.
Answer:
A bidentate ligand like (en), can form two coordinate bonds with the metal at the same time.
An amidentate ligand like -NO₂ can form only one coordinate bond with the metal at a time. But it can ligate through two different atoms.

Plus Two Chemistry Coordination Compounds Four Mark Questions and Answers

Question 1.
Match the following table:

Answer:

Question 2.
1. Write the IUPAC names of the following coordination compounds.

• [Pt(NH₃)Cl(NO₂]₂)
• K₃[Cr(C₂O₄)₃]

2. Identify the type of isomerism exhibited by the following complexes and distinguish them. [Co(NH₃)₅SO₄] Br and [Co(NH₃)₅Br] SO₄
Answer:
1. The IUPAC names of the coordination compounds:

• Amminechloridonitrito-N platinum(II)
• Potassiumtrisoxalatochromate(III)

2. Ionisation isomerism.
The first compound gives pale yellow ppt. with AgNO₃ solution whereas the second compound gives white ppt. with BaCl₂ solution.

Question 3.
1. Write the IUPAC name of the following compounds:

• [Pt(NH₃)₂Cl₂]
• K₄[Fe(CN)₆]

2. A list of coordination compounds are given below:

• [Cr(H₂O)₆]Cl₃,
• [CO(NH₃)₅ Br]SO₄,
• [CO(NH₃)₅ NO₂]²⁺,
• [CO(NH₃)₆] [Cr(CN)₆]

Which type of isomerism do these compounds exhibit?
Answer:
1. The IUPAC name of the coordination compounds:

• Diamminedichloridoplatinum(II)
• Potassium hexacyanoferrate(II)

2. Type of isomerism

• [Cr(H₂O)₆]Cl₃ – Hydrate isomerism
• [CO(NH₃)₅ Br]SO₄ – Ionisation isomerism
• [CO(NH₃)₅ NO₂]²⁺ – Linkage isomerism
• [CO(NH₃)₆] [Cr(CN)₆] – Coordination isomerism

Question 4.
1.. Write down the IUPAC name of

• K₄[Fe(CN)₆]
• [Pt(NH₃)₂Cl₂]

2. On the basis of VBT, explain why [Fe(H₂O)₆]²⁺ is strongly paramagnetic while [Fe(CN)₆]^3- is weakly paramagnetic.
Answer:
1. The IUPAC names are:

• Potassium hexacyanoferrate(II)
• Diamminedichloridoplatinum(II)

2.

In [Fe(H₂0)₆]^2-, iron undergoes sp³d² hybridisation (inner orbital complex). It has four unpaired electrons and hence it is highly paramagnetic.

In [Fe(CN)₆]³⁺. iron undergoes d²sp³ hybridisation (outer orbital complex). It has only one unpaired electron. Hence, it is less paramagnetic.

Question 5.
The names of some co-ordination compounds are given below:

• EDTA
• Haemoglobin
• cis-platin
• Vitamin B¹²
• D-penicillamine
• Chlorophyll
• Ni(CO)₄

a. Classify the above compounds on the basis of application of coordination compounds?
b. There are given some of the coordination compounds Name them.

1. K₃[Fe(C₂O₄)₃]
2. [Cr(CN)₃]³⁺
3. [CoSO₄(NH₃)₄]NO₃
4. [CO(NO₂)₃(NH₃)₃]

Answer:
a. On the basis of application of coordination compounds:

• In biological system – Haemoglobin, Vitamin B₁₂, Chlorophyll.
• Estimation of hardness of water – EDTA.
• Extraction of metals – Ni(CO)₄
• In medicine – D-penicillamine, cis-platin

b. The coordination compounds are:

1. K₃[Fe(C₂O₄)₃] – Potassiumtrioxalatoferrate(III)
2. [Cr(CN)₃]³⁺ – Trisethylenediaminechromium(III) ion
3. [CoSO₄(NH₃)₄]NO₃ – Tetraamminesulphato- cobalt(III) nitrate
4. [CO(NO₂)₃(NH₃)₃] – Triamminetrinitrito-N-cobalt(III)

Question 6.
1. Name the following compounds.

• [Pt(NH₃)₄ ] [CuCl₄]
• [PtCl₂ (NH₃)₄] Br₂

2. What type of isomerism is shown by the following coordination compounds?

• [Pt(NH₃)₄ ] [CuCl₄]
• [Cr(en)₃]³⁺

Answer:
1.

• Tetraammineplatinum(II) tetrachlorocuprate(II)
• Tetraamminedichloridoplatinum(IV) bromide

2. Type of isomerism:

• Coordination isomerism
• Optical isomerism

Question 7.

1. Write the d-orbital configuration of [Ti(H₂O₆]²⁺
2. Ti⁴⁺ is colourless. Why?
3. Write the possible isomers of [Co(NH₃)₅ Br] SO₄ and name them.

Answer:
1.

2. In Ti⁴⁺ there are no electrons in 3d orbitals. Hence ‘d-d’ transition cannot take place. Hence it is colourless.

3. Possible isomers of [Co(NH₃)₅ Br] SO₄ and names

• [CO(NH₃)₅ Br] SO₄ – Pentamminebromido- cobalt(III) sulphate
• [Co(NH3)5SO4]Br-Pentamminesulphatecobalt(III) bromide

Question 8.
‘A’ and ‘B’ are isomers. They have the same composition. But ‘B’ cannot give the test for sulphate.

1. Write two suitable coordination compounds which give the test for sulphate.
2. What are the two major classes of isomerism exhibited by coordination compounds?
3. Draw the structure of an octahedral complex that show optical isomerism.

Answer:
1. Two suitable coordination compounds which give the test for sulphate

• [CO(NH₃)₅Cl]SO₄
• [Co(NH₃)₅Br] SO₄

2. Structural isomerism, Stereoisomerism
3. [PtCl₂(en)₂]²⁺

Question 9.

1. Write the formula of the complex ion Chloridonitro tetramminecobalt(III)?
2. Identify the ligands, coordination number and coordination sphere.
3. Explain the structure of Tetracarbonylnickel(O) with the help of Valence Bond Theory.

Answer:

1. [CO(NH₃)₄Cl(NO₂^–)
2. Ligands-NH₃, Cl^– , NO₂^– Coordination number -4
3. Tetracarbonylnickel(0) – [Ni(CO)]₄] – Structure

Ni(CO₄) is diamagnetic as it does not contain any unpaired electron.

Question 10.
Consider the complex ion [Ti (H₂O)₆]³⁺

1. What is its outer electronic configuration and its shape?
2. What do you mean by crystal field splitting theory?

Answer:

1. Ti³⁺ – 3d¹ 4s⁰ Octahedral
2. In the case of an isolated gasesous metal atom/ ion all the five d-orbitals have the same energy (degenerate). Due to the presence of ligands are splitted the degeneracy of the d-orbitals is lifted.

Question 11.
Coordination compounds are those compounds which retain their identity even in solution and it is essential for all living matter.

1. Name a coordination compound containing Magnesium, which is essential for plants.
2. When we coordinate EDTAwith any metal, we get a ring structure. What is this process called?
3. Explain.

Answer:

1. Chlorophyll.
2. Chelation.
3. When a poly dentate ligand is coordinated to the metal, a ring structure is obtained, called chelate complex. Such complexes are more stable than similar complexes containing unidentate ligands.

Question 12.
Some ligands are given below. Arrange them in suitable headings.
[H₂O, NH₃, CN^–, CO, Cl^–, OH^– (en)]
1. What do you mean by the term ligand?
2. Write down the nomenclature of the coordination compounds given below.

• K₄[Fe(CN)₆]
• [Ag (NH₃)₂]Cl

Answer:
Neutral ligands – H₂O, NH₃, CO, en
Charged ligands – CN^–, OH^–, Cl^–
1. Ligand is a neutral molecule or charged ion which can donate atleast one lone pair of electron to the metal.
2. nomenclature of the coordination compounds

• Potassium hexacyanoferrate(II)
• Diamminesilver(I) chloride

Question 13.

1. What do you mean by optically active compounds? Give two examples.
2. Draw the ‘d’ and T forms of [Co(en)₃]³⁺.

Answer:
1. Optically active compounds are formed by chiral moneluces i.e., molecules which do not have plane of symmetry. These isomers are non- superimposable mirror images of each other. They are optically active and rotate the plane of polarised light equally but in opposite directions.

The isomer which rotates the plane of polarised light towards left is called leavorotatory (-) while that which rotate plane towards right is called dextrorotatory (+).
e.g. [Co(en)₃]³⁺, [PtCl₂(en)₂]²⁺
Dextro and laevo forms of these compounds are possible.

2.

Question 14.
What is the importance of the following coordination compounds is different fields?

1. EDTA
2. Gold cyanide [AU(CN)₂]
3. cis-platin
4. [Ag(S₂O₃)]^3-

Answer:

1. EDTA → Estimation of hardness of water
2. AU(CN)₂ → Metallurgy
3. cis-platin → Cancer therapy
4. [Ag(S₂O₃)]^3- → Photography

Question 15.
The d-block elemetns forms coordination compounds.

1. Name the coordination compound K₃ [CoF₆].
2. Write the electronic configuration of the central metal atom of the above complex by using CFT
3. Draw the figure to show the splitting of degenerate, ‘d’ orbitals in an octahedral field.

Answer:
1. K₃[CoF₆] – Potassiumhexafluridocobaltate(III)
2. The electronic configuration of Co (Z = 27) is [Ar]3d⁷4s² In K₃ [CoF₆], Co is in +3 state. The configuration of Co³⁺ is 3d⁶ 4s°.
3.

Question 16.

1. Name the compound K₃[Cr(C₂O₄)₃]
2. Explain on the basis of VB theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and [Ni(Cl)₄]^2- ion with tetrahedral structure is paramagnetic.

Answer:

1. K₃[Cr(C₂O₄)₃] – Potassiumtrioxalatochromate(III)
2. Ni = 1s²2s²2p63s²3p64s²3d²

It is diamagnetic due to absence of unpaired electrons.

∴ Due to presence of unpaired electrons it is paramagnetic.

Plus Two Chemistry Coordination Compounds NCERT Questions and Answers

Question 1.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu²⁺ ion. Explain why?
Answer:
FeSO₄ does not form any complex with (NH₄)₂SO₄. Instead, it forms a double salt FeSO₄.(NH₄)₂SO₄.6H₂O which dissociates completely into ions. CuSO₄ when mixed with NH₃ forms a complex [CU(NH₃)₄]SO₄ in which the complex ion [CU(NH₃)₄]²⁺ does not dissociate to give Cu²⁺ ion.

Question 2.
Write the geometrical isomers of [Pt(NH₃)(Br)(Cl) (py)] and how many of these will exhibit optical isomerism?
Answer:
The complex [Pt(NH₃)(Br)(Cl)(py)] will form three geometrical isomers:

Square planar complexes of this type will not show geometrical isomerism.

Question 3.
Aqueous copper sulphate solution (blue in colour) gives:

1. a green precipitate with aqueous potassium fluoride and
2. a bright green solution with aqueous potassium chloride. Explain these experimental results.

Answer:
Aqueous copper sulphate contains coordination entity, [Cu(H₂O)₄]²⁺ which is blue in colour. Water molecule is a weaker ligand than Cl^– and F^–.
1. On addition of aqueous KF solution, a new complex entity is formed wich is green in colour.

2. On addition of aqueous solution of KCl, another bright green complex entity is formed which is soluble in water.

Question 4.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S(g) is passed through this solution?
Answer:
Aqueous solution of copper sulphate contains Cu²⁺ ions in the form of complex entity, [Cu(H₂O)₄]²⁺ and H₂O ligand is a weak ligand. When excess of KCN is added, a new coordination entity, [Cu(CN)₄]^2- is formed due to the following reaction:

Cyanide ligand, CN^– is a strong field lignad and stability constant of [Cu(CN)₄]^2- is quite large and thus practically no Cu²⁺ ions are left in solution. On passing H₂S gas, no CuS is formed due to non-availabiliy of Cu²⁺ ions in solution.

Question 5.
What will be the correct order for the wavelengths of absorption in the visible region for the following:
[Ni(NO₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺
Answer:
The order of the ligand in the spectrochemical series is H₂O < NH₃ < NO₂^–. Hence the wavelength of the light observed will be in the order:
[Ni(H₂O)₆]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(NO₂)₆]^4-

Thus, the wavelength of light absorbed (E = \(\frac{\mathrm{hc}}{\lambda}\)) will be in the opposite order:
[Ni(H₂O)₆]²⁺ > [Ni(NH₃)₆]²⁺ > [Ni(NO₂)₆]^4-

Students can Download Chapter 9 Ray Optics and Optical Instruments Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 9 Ray Optics and Optical Instruments

Plus Two Physics Ray Optics and Optical Instruments NCERT Text Book Questions and Answers

Question 1.
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is

1. a convex lens of focal length 20 cm, and
2. a concave lens of focal length 16 cm?

Answer:
Here the object is virtual and the image is real.
u = 12cm object on right and virtual.
1. f = +20 cm

i.e., v = 7.5 cm. (image on right and real). It is located 7.5 cm from the lens.

2. f = -16 cm

i.e., v = 48 cm. (image on right and real). Image will be located 48 cm from the lens.

Question 2.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length of the lens is to be 20 cm?
Answer:
\(\frac{\mu_{2}}{\mu_{1}}\) = µ = 1.55
R₁ = R₂ = R
f = 20 cm

R = 0.55 × 2 × 20 = 22 cm.

Question 3.
A small telescope has an objective lens of focal length 144 cm and eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Answer:
1. For normal adjustment.
M.P. of telescope = \(\frac{f_{0}}{f_{e}}=\frac{144}{6}\) = 24

2. The length of the telescope in normal adjustment
L = f_o + f_e
= 144 + 6 = 150 cm.

Plus Two Physics Ray Optics and Optical Instruments One Mark Questions and Answers

Question 1.
Fora total internal reflection, which of the following is correct?
(a) Light travel from rarer to denser medium.
(b) Light travel from denser to rarer medium.
(c) Light travels in air only.
(d) Light travels in water only.
Answer:
(b) Light travel from denser to rarer medium.
Explanation: In total internal reflection, light travel from denser to rarer medium.

Question 2.
Focal length of a convex lens of refraction index 1.5 is 2 cm. The focal length of lens, when immersed in a liquid of refractive index of 1.25, will be.
(a) 10 cm
(b) 2.5 cm
(c) 5 c
(d) 7.5 cm
Answer:
(c) 5 c

Question 3.
If the refractive index of a material of equilateral prism is \(\sqrt{3}\), then angle of minimum deviation of the prism is
(a) 60°
(b) 45°
(c) 30°
(d) 75°
Answer:
(a) 60°
Explanation: A = 60°, n = \(\sqrt{3}\), D = ?

D = 60°.

Question 4.
Which of the following is correct for the beam which enters the medium?
(a) Travel as a cylindrical beam
(b) Diverge
(c) Converge
(d) Diverge near the axis and converge near the periphery
Answer:
(c) Converge.
Explanation: Since the refractive index is less at the beam boundary, the ray at the edges of the beam move faster compared to the axis of beam. Hence, the beam converges.

Question 5.
A beam of monochromatic light is refracted from vacuum into a medium of refraction index 1.5. the wavelength of refracted light will be.
(a) Depend on intensity of refracted light
(b) Same
(c) smaller
(d) larger
Answer:
(c) smaller
Explanation: velocity of light decreases in a medium. Hence λ decrease in a medium (v ∝ λ).

Question 6.
A convex lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a combination of lenses. What is the power in diopters of the combination is
Answer:
Focal length of convex lens f₁ = 25 cm
Focal length of concave lens f₂ = -25 cm
Power of combination in diopters,

Question 7.
Fill in the blanks

Answer:
(i) n = \(\frac{c}{v}=\frac{3 \times 10^{8}}{2 \times 10^{8}}\) = 1.5

(ii) Optical fibre

Plus Two Physics Ray Optics and Optical Instruments Two Mark Questions and Answers

Question 1.
Match the following.

A	B
Double convex	(R1 -ve, R2 +ve)
Double concave	(R, = ∞, R2 +ve)
Plane convex	(R1 +ve, R2-ve)
Plane concave	(R = ∞, R2 -ve)

Answer:

A	B
Double convex	(R, +ve, R2-ve)
Double concave	(R1 -ve, R2 +ve)
Plane convex	(R1 = α, R2 -ve)
Plane concave	(R1 = α, R2+ve)

Plus Two Physics Ray Optics and Optical Instruments Three Mark Questions and Answers

Question 1.
A hemispherical transparent paperweight of radius 5m and refractive index 1.5 is placed on a table. A beam of lazar, at a distance of 2m from the centre is directed as shown in the figure.
1. Name the law which is related to refraction.
2. Locate the position of the image by completing the ray diagram.

3. With the source of laser at the centre of the hemisphere, redraw the ray diagram.
Answer:
1. Snell’s law
2.

3. The refracted ray is undeviated.

Question 2.
Figure (a) below snows the image observed at the near point of eye by a boy through a simple microscope. Eye focused on near point.

1. Draw ray diagram which shows the image formation at infinity, so that the boy can observe it with a relaxed eye.
2. Distinguish between linear magnification and angular magnification.

Answer:
1.

2. Linear magnification is ratio of image height to object height. Angular magnifications is the ratio of angle subtended by the image and the object on the eye when both are at the least distance of distinct vision.

Question 3.
Figure shows the path of the light rays through a glass slab.

1. Name the phenomena involved here.
2. Relate the values of n_1, n₂, i and r on the basis of one figure.
3. Copy the figure of glass and draw the path of ray when n₂ < n₁.

Answer:
1. Refraction

2. \(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)

3.

Question 4.
A light ray travelling from one medium to another medium is given in the figure.

1. Write a mathematical relation for this refraction.

• n₂ < n₁
• n₂ > n₁
• n₂ = n₁

2. What is a relation between angle of incidence, angle of refraction and refractive index of medium.

3. A flint glass rod when immersed in carbon disulfide is nearly invisible why?

Answer:
1. n₂ < n₁

2. \(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)

3. Refractive index of flint glass rod and carbon disulfide are nearly equal. Hence no refraction (or reflection) take place.

Question 5.
A convex lens and concave lens are placed as shown in figure. For convex lens f = 10cm for concave it is 5 cm

1. Is it converging or diverging why?
2. If f₁ = 5cm and f₂ =10cm What change will occur in the optical nature of system?

Answer:
1.

f = -10 cm
Effective focal length is negative. Hence this lens is diverging.

2. Effective focal length becomes positive- Hence the lens will act as converging.

Plus Two Physics Ray Optics and Optical Instruments Four Mark Questions and Answers

Question 1.
The maximum possible magnification for a simple microscope is 10

1. How do you increase the magnification further(1)
2. Draw the ray diagram for compound microscope and find an expression for magnification (3)
3. What is the advantage of forming image at infinity? (1)

Answer:
1. Use two convex lens instead of single lens.

2.

The magnification produced by the compound microscope

Multiplying and dividing by I₁M₁ we get,

Where m₀ & m_e are the magnifying power of objective lens and eyepiece lens.
∴ m = m_e × m₀ ______(1)
Eyepiece acts as a simple microscope.
Therefore \(\mathrm{m}_{\mathrm{e}}=1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\) ______(2)
We know magnification of objective lens
m₀ = \(\frac{V_{0}}{u_{0}}\) ______(3)
Where v₀ and u₀ are the distance of the image and object from the objective lens.
Substituting (2) and (3) in (1), we get

for compound microscope, u₀ » f₀ (because the object of is placed very close to the principal focus of the objective) and v₀ ≈ L, length of microscope (because the first image is formed very close to the eye piece).

where L is the length of microscope, f₀ is the focal length of objective lens.

3. Strain for eye, will be minimum when image is at infinity.

Question 2.
The refraction of light travelling from glass to water is shown in the figure.
1. The snells law in the above case can be written as………..

2. Show that c = \(\sin ^{-1}\left(_{g} n_{w}\right)\). Where C is the critical angle of glass water interface. (2)

3. Three light rays, (Red, blue and yellow) incident at one side and its refractions are shown in the figure. Copy the figure and mark Red, blue and yellow in the figure. (1)

Answer:
1.

2. In this i = c using snell law, we can write

3.

Question 3.
When a point object is placed in front of a spherical refracting surface an image is formed in the refracting medium.
1.

Complete the ray diagram to locate the position of the image.

2. Obtain the expression \(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R}\) for the position of image inside refracting medium.

3. If the refracting surface is concave in nature, with the same set up, locate the position of the image by drawing a ray diagram.
Answer:
1.

2. Refraction at a spherical surface:

Consider a convex surface XY, which separates two media having refractive indices n₁ and n₂. Let C be the centre of curvature and P be the pole. Let an object is placed at ‘O’, at a distance ‘u’ from the pole.

I is the real image of the object at a distance V from the surface. OA is the incident ray at angle ‘i’ and Al is the refracted ray at an angle ‘r’. OP is the ray incident normally. So it passes without any deviation. From snell’s law,

\(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)
If ‘i’ and ‘r’ are small, then sin i » i and sin r » r.

n₁i = n₂r ……….(1)
From the Δ OAC, exterior angle = sum of the interior opposite angles
i-e., i = α + θ ………(2)
Similarly, from Δ IAC,
a = r + β
r = α – β ……..(3)
Substituting the values of eq(2) and eq(3)in eqn.(1) we get,
n₁(α + θ) = n₂(α – β)
n₁α + n₁β = n₂α – n₂β
n₁θ + n₂β = n₂α – n₁α
n₁θ + n₂β = (n₂ – n₁)α ………….(4)
From OAP, we can write,

From IAP, β = \(\frac{\mathrm{AP}}{\mathrm{PI}}\), From CAP, α = \(\frac{\mathrm{AP}}{\mathrm{PC}}\)
Substituting θ, β and α in equation (4) we get,

According to New Cartesian sign convection, we can write,
OP=-u, PI = +v and PC = R
Substituting these values, we get


Case -1:
If the first medium is air, n₁ = 1, and n₂ = n,

3.

Question 4.
A compound microscope consists of an objective lens of focal length 2cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at

1. Least distance of distinct vision.
2. infinity

Answer:
1. v_e = -25cm

∴ u_o = 5cm
Length of the tube, L= |v_o| + |u_e|
∴ v_o = 15-5 = 10

u_e = -2.5cm

2. ∴ v_o = 15 – f_e = 15 – 6.25 = 8.75

u_o = -2.59cm

Question 5.
You may be observed that, the fish inside the aquarium appears to be raised.
1. What is the reason for this phenomenon?
2. Obtain an expression for apparent shift offish.
3. What happens to the height of the object, (That vertically stands in the aquarium) when it is observed by the fish.

• becomes taller
• becomes smaller
• Does not change the height. Justify your answer.

Answer:

1. Refraction
2. Expression for apparent shift is not included in the syllabus
3. Becomes taller. When light enters from rare to denser medium, it deviates towards the normal.

Question 6.
High precision optical instruments uses prisms instead of mirror to reflect light.

1. Name the phenomena used for reflecting light using prism.
2. What is the advantage of using prism instead of mirror for reflecting light?
3. The critical angle of water is 52°. Calculate the refractive index of water.

Answer:
1. Total internal reflection

2. Prism can be used for total internal reflection. Mirrors can’t be used for total internal reflection.

3.

n = 1.26.

Question 7.
Two lenses of focal lengths f₁ and f₂ are placed in contact

1. If the object is at principal axis, draw ray diagram of the image formation by this lens.
2. Obtain a general expression for effective focal length in terms of f₁ and f₂.
3. How will you combine a convex lens of focal length f₁ and concave lens f₂ such that combination acts as

• Converging
• diverging
• plane glass plate

Answer:
1.

2. Obtain an expression for the effective focal length of the combination of two thin convex lenses in contact.

3.

• Keep in a medium of refractive index lower than that of lens.
• Keep in a medium of refractive index higher than that of lens.
• Keep in a medium of refractive index equal to refractive index of lens.

Question 8.
The light rays travelling from rarer to denser medium is given in the figure
1. Redraw the diagram and correct it

2. State the law relating i and r for retracted ray.
3. Velocity of light in water is 2.25 × 10⁸ m/s, If angle of incidence is 30° calculate angle of refraction.
Answer:
1.

2. Snells law:
The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media and for the given colour of light used. This constant is known as the refractive index of second medium w.r. t. the first medium.

Explanation: If ‘i’ is the angle of incidence in the first medium and ‘r’ is the angle of refraction in the second medium, then by Snell’s law,

Where ₁n₂ is the refractive index of the second medium with respect to the first medium. If the first medium is air, then sin i/sin r is known as absolute refractive index of the second medium.

where ‘n’ is the refractive index of the second medium.

3.

Question 9.

1. An air bubble inside an ice block shine brilliant by……… (Refraction, Reflection, total internal reflection)
2. Explain the above phenomenon.
3. The light ray incident at one face of the prism is shown in figure. Copy this figure complete the path of the ray. (Take critical angle of prism C = 42°)

Answer:
1. Total internal reflection.

2. Whenarayoflightpassesfromadenserto rarer medium, after refraction the ray bends away from the normal. If we increases the angle of incidence beyond the critical angle, the ray is totally reflected back to the denser medium itself. This phenomenon is called total internal reflection.

3.

Question 10.
A convex lens produces an inverted image of size 1.4cm The size of object is 0.7cm

1. What is magnification in the case
2. What is the nature of image
3. If the object is at distance 30 cm from the lens calculate focal length of the lens

Answer:
1.

2. Real, inverted, magnified

3.

Question 11.
A ray of light parallel to the principal axis of a spherical mirror falls at a point M as shown in the figure.

1. Identify the type of the mirror used in the diagram. (1)
2. If focal length of the mirror is 10cm, what is the distance CF in the figure? (1)
3. Complete the ray diagram and mark the angle of incidence and angle of reflection. (2)
4. If the mirror is immersed in water its focal length will be

• less than 10cm
• 10cm
• greater than 10cm
• 20cm

Answer:
1. Concave mirror.

2. 10 cm (CF = PC – PF = R = f – 2f – f = f = 10cm).

3.

4. 10cm. Focal length of mirror is independent of medium.

Question 12.
The given figure shows a compound microscope with two lenses PQ and RS.

1. Identify Objective and eyepiece in the microscope.
2. A compound microscope has a magnification of 30. The focal length of its eyepiece is 5cm. Assuming the final image to be formed at the least distance of distinctive vision, calculate the magnification produced by the objective.
3. What is the length of a compound microscope in normal adjustment?

Answer:
1. Objective – PQ, eyepiece – RS.
2. Magnification, M = m₀ × m_e

3. The length of a compound microscope in normal adjustment is f₀ + f_e.

Question 13.
A compound microscope consists of an objective lens of focal length 2cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at

1. Least distance of distinct vision.
2. infinity

Answer:
1. v_e = -25cm

∴ u₀ = 5cm
Length of the tube, L= |v₀| + |u_e|
∴ v₀ = 15 – 5 = 10

u_e = -2.5 cm.

2. ∴ v₀ = 15 – f_e = 15 – 6.25 = 8.75

u₀ = -2.59cm

Plus Two Physics Ray Optics and Optical Instruments Five Mark Questions and Answers

Question 1.
A point object at a distance of 36 cm from the convex lens of focal length 10cm, is moved by 10cm in 2 sec along principle axis towards the lens. Then image will also change its position.

1. Write the law which relates object and image distance from the lens.
2. Find the initial and final position of the image and calculate average speed of image.
3. A man argues that the image will move uniformly at the same speed as that of object. What is your opinion? Justify.

Answer:
1. The lens equation \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

2. u = -36, f = 10
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

If object is moved 10cm towards the lens we can find position
u = -26, f=10
v = 7.2 cm
Speed = \(\frac{7.8-7.2}{2}\) = 3cm/sec.

3. Comparing speed of object and image we can arrive at conclusion that the argument of man is false, speed of image is different from speed of object.

Question 2.
A ray of light parallel to the principal axis of a spherical mirror falls at a point M as shown in the figure.

1. Identify the type of the mirror used in the diagram.
2. If focal length of the mirror is 10cm, what is the distance CF in the figure?
3. Complete the ray diagram and mark the angle of incidence and angle of reflection.
4. If the mirror is immersed in water its focal length will be

• less than 10cm
• 10cm
• greater than 10cm
• 20cm

Answer:
1. Concave mirror.

2. 10 cm (CF = PC – PF = R = f – 2f – f = f = 10cm).

3.

4. 10cm. Focal length of mirror is independent of medium.

Question 3.
The image formed by a thin lens is shown in the figure.

1. What is the nature of the image
2. Find out the power of the image
3. Draw the ray diagram showing above lens forming a magnified erect, virtual image
4. If a convex lens of focal length 20cm is kept in contact with above lens. What is the focal length and power of the combination

Answer:
1. Inverted.

2. P= \(\frac{1}{1}\) =ID.

3.

4. P = P₁ + P₂
in this case f₁ = 1 m, f₂ = 0.2m

Question 4.
A beam of light passing from one transparent medium to another obliquely, undergoes an abrupt change in direction. This phenomenon is known as refraction of light.

1. Name the law which satisfies during this refraction.
2. Draw a figure, which shows refraction through a parallel sided glass slab (Ray passing from air)
3. Using the figure obtained in (b), show that the incident ray and the emergent ray are parallel to each other. Redraw the same figure, if the light is entering from a medium denser than glass. Justify your answer.

Answer:
1. Snell’s law.

2.

3. This derivation is out of syllabus.

4. The light bends away from the normal if light enter from glass to water.

Question 5.
A group of students are given a project for constructing a telescope and they were supplied with two biconvex lens of power 1 diopter and 0.1 dioptre.

1. Of the two lenses, which can be used as objective?
2. Draw the ray diagram for the formation of the image by a telescope.
3. Arrive at an expression for magnification of a telescope.
4. Prepare a notice/ label about the precaution to be taken while using the telescope and limitations of the telescope constructed.

Answer:
1. Biconvex lens of power 0.1 dioptre.

2.

3. Magnification:
The magnifying power of a telescope is the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the objective.
\(\mathrm{m}=\frac{\text { angle subtented by the image at eye (eye piece) }}{\text { angle subtended by the object at the objective }}\)
i.e. m = \(\frac{β}{α}\) …….(1) [from figure]
But from ∆CIM, tanα = \(\frac{IM}{IC}\), α = \(\frac{IM}{IC}\)
(For small values tan α ≈ α)
from ∆C¹IM, tanβ = \(\frac{IM }{\mathrm{IC}^{1}}\) , β = \(\frac{IM }{\mathrm{IC}^{1}}\)
substituting α and β in eq (1) we get

But IC = f_o (the focal length objective lens) and IC^l = f_e (the focal length eyepiece lens.)

In this case the length of the telescope tube is (f₀ + f_e).
Case 1 :
When the image formed by the objective is within the focal length of the eyepiece, Then the final image is formed at the least distant of distinct vision. In this case, magnifying power.

4.

• As magnifying power is negative, the final image in an astronomical telescope is inverted.
• To have large magnifying power, f_o must be as large as possible and f_e must be as small as possible.
• As intermediate image is between the two lens, cross wire (ora measuring device) can be used.
• In normal setting of telescope, the final image is at infinity.

Question 6.
The following graph represent id curve of a optical instrument placed in air.

1. Name the device which give the above i-d curve.
2. Obtain an expression for deviation produced by such a device.
3. What is the relevance of the value ‘D’? Arrive at an expression for refractive index in terms of this value from (b).
4. How is the deviation affected if the above arrangement is immersed in a liquid of refractive index less than that of the above device.

Answer:
1. Prism.

2. Refraction through a prism:

ABC is a section of a prism. AB and AC are the refracting faces, BC is the base of the prism. ∠A is the angle of prism. Aray PQ incidents on the face AB at an angle i₁. QR is the refracted ray inside the prism, which makes two angles r₁ and r₂ (inside the prism). RS is the emergent ray at angle i₂.

The angle between the emergent ray and incident ray is the deviation ‘d’.
In the quadrilateral AQMR,
∠A + ∠R = 180°
[since N₁M are normal]
ie, ∠A + ∠M = 180°____(1)
In the ∆ QMR,
∴ r₁ + r₂ + ∠M = 180° ____(2)
Comparing eq (1) and eq (2)
r₁ + r₂ = ∠A ____(3)
From the ∆ QRT,
(i₁ – r₁) + (i₂ – r₂) = d
[since exterior angle equal sum of the opposite interior angles]
(i₁ + i₂) – (r₁ + r₂) = d
but, r₁ + r₂ = A
∴ (i₁ + i₂) – A = d

3. At minimum deviation D = 2i – A, r₁ = r₂ = r

4. Deviation decreases.

Question 7.
\(\frac{1}{f}=\left(\frac{n_{2}}{n_{i}}-t\right)\left(\frac{1}{R_{i}}-\frac{1}{R_{2}}\right)\) is lens maker’s formula.
1. Write down lens maker’s formula for a convex lens.
2. “If a convex lens is immersed in water its converging power decrease. Do you agree with it? Justify your answer.
3. A convex lens of refractive index n₂ is placed in different media. Explain optic behavior in each. If n₁ is refractive index of surrounding media.

• in medium with n₂>n₁
• in a medium with n₂ < n₁
• in a medium n₁ = n₂

Answer:
1. For convex lens R₁ = +ve and R₂ = -ve

2. Yes.

From the above equation it is clear that, \(\mathrm{P} \alpha \frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\)

In water, \(\frac{n_{2}}{n_{1}}\) is less. Hence power decreases.

3. P = +ve, converging
P = -ve, diverging
P = 0, Plane glass

Question 8.
Two convex lense are given in the figure A and figure B

1. Which has more curvature
2. Which has more power
3. Which lens produce more magnification
4. which lens has less focal length
5. Can these lenses act as diverging lenses in any condition?

Answer:

1. A
2. A
3. A
4. A
5. Yes, If we place this lens in a medium of higher refractive index than lens.

Question 9.
In the figure given below, PQ represents an incident ray falling in the side AB of a prism, when monochromatic light is used

1. Draw the refracted ray, emergant ray and mark the angle of deviation
2. Derive an equation for refractive index of the material of the prism in terms of angle of minimum deviation
3. Draw the incident ray and refracted ray, at the angle of minimum deviation

Answer:
1.

2. Refraction through a prism:

ABC is a section of a prism. AB and AC are the refracting faces, BC is the base of the prism. ∠A is the angle of prism. Aray PQ incidents on the faceAB at an angle i₁. QR is the refracted ray inside the prism, which makes two angles r₁ and r₂ (inside the prism). RS is the emergent ray at angle i₂.

The angle between the emergent ray and incident ray is the deviation ‘d’.
In the quadrilateral AQMR,
∠A + ∠R = 180°
[since N₁M are normal]
ie, ∠A + ∠M = 180°____(1)
In the ∆ QMR,
∴ r₁ + r₂ + ∠M = 180° ____(2)
Comparing eq (1) and eq (2)
r₁ + r₂ = ∠A ____(3)
From the ∆ QRT,
(i₁ – r₁) + (i₂ – r₂) = d
[since exterior angle equal sum of the opposite interior angles]
(i₁ + i₂) – (r₁ + r₂) = d
but, r₁ + r₂ = A
∴ (i₁ + i₂) – A = d
(i₁ + i₂) = d + A ____(4)
It is found that for a particular angle of incidence, the deviation is found to be minimum value ‘D’.
At the minimum deviation position,
i₁ = i₂ =i, r₁ = r₂ = r and d = D
Hence eq (3) can be written as,
r + r = A
or r = \(\frac{A}{2}\) ____(5)
Similarly eq (4) can be written as,
i + i = A + D
n = \(\frac{A + D}{2}\) ____(6)
Let n be the refractive index of the prism, then we can write,
n = \(\frac{sin i}{sin r}\) ____(7)
Substituting eq (5) and eq (6) in eq (7),
\(n=\frac{\sin \frac{A+D}{2}}{\sin \frac{A}{2}}\)
i – d curve:

It is found that when the angle of incidence increases deviation (d) decreases and reaches a minimum value and then increases. This minimum value of the angle of deviation is called the angle of minimum deviation.

3. Refracted ray is parallel to base

Question 10.
Refraction of a ray of light at a spherical surface separating two media having refractive indices n₁ and n₂ is shown in the figure.

1. Which of the two media is more denser?
2. In the figure, show that\(\frac{\mathrm{n}_{1}}{\mathrm{OA}}+\frac{\mathrm{n}_{2}}{\mathrm{AI}}=\frac{\mathrm{n}_{2}-\mathrm{n}_{1}}{\mathrm{AC}}\).
3. Using the above relation arrive at the thin lens formula.
4. An object is placed on the principal axis of a convex lens at a distance 8 cm from it. Find the magnification of the image if the focal length of the lens is 4 cm.

Answer:
1. Refractive index of medium 2 is greater than medium 1.

2. Refraction at a spherical surface

Consider a convex surface XY, which separates two media having refractive indices n₁ and n₂. Let C be the centre of curvature and P be the pole. Let an object is placed at ‘O’, at a distance ‘u’ from the pole.

I is the real image of the object at a distance V from the surface. OA is the incident ray at angle ‘i’ and Al is the refracted ray at an angle ‘r’. OP is the ray incident normally. So it passes without any deviation. From snell’s law,

\(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)
If ‘i’ and ‘r’ are small, then sin i » i and sin r » r.

n₁i = n₂r ……….(1)
From the Δ OAC, exterior angle = sum of the interior opposite angles
i-e., i = α + θ ………(2)
Similarly, from Δ IAC,
α = r + β
r = α – β……..(3)
Substituting the values of eq(2) and eq(3)in eqn. (1) we get,
n₁(α + θ) = n₂(α – β)
n₁α + n₁β = n₂α – n₂β
n₁θ + n₂β = n₂α – n₁α
n₁θ + n₂β = (n₂ – n₁)α ………….(4)
From OAP, we can write,

From IAP, β = \(\frac{\mathrm{AP}}{\mathrm{PI}}\), From CAP, a= \(\frac{\mathrm{AP}}{\mathrm{PC}}\)
Substituting θ, β, and α in equation (4) we get,

According to New Cartesian sign convection, we can write,
OP=-u, PI = +v and PC = R
Substituting these values, we get


Case -1:
If the first medium is air, n₁ = 1, and n₂ = n,

3. Consider a thin lens of refractive index n₂ formed by the spherical surfaces ABC and ADC. Let the lens is kept in a medium of refractive index n₁. Let an object ‘O’ is placed in the medium of refractive index n₁ Hence the incident ray OM is in the medium of refractive index n₁ and the refracted ray MN is in the medium of refractive index n₁.

The spherical surface ABC (radius of curvature R₁) forms the image at I₁ Let ‘u’ be the object distance and ‘v₁’ be the image distance.
Then we can write,

This image I₁ will act as the virtual object for the surface ADC and forms the image at v.
Then we can write,

Dividing throughout by n₁, we get

if the lens is kept in air, \(\frac{n_{2}}{n_{1}}\) = n
So the above equation can be written as,

From the definition of the lens, we can take, when
U = ∞, f = v
Substituting these values in the eq (3), we get

This is lens maker’s formula

For convex lens,
f = +ve, R₁ = +ve, R₂ = – ve

For concave lens,
f = -ve, R₁ = -ve, R₂ = +ve

Lens formula From eq(4),

4. u = -8cm, f = +4cm
m = \(\frac{f}{f+u}=\frac{4}{4+-8}\)
m = -1.

Question 11.
Two lenses L₁ and L₂ are placed in contact as shown in figures. The focal length of each lens is 10cm

1. What is power of L₁
2. What is power of L₂
3. What is effective focal length of combination
4. “The power of convex is greater than that of concave and combination can act as a diverging lens”. Is this statement true in any situation? Explain?

Answer:
1.

2.

3.

The combination will act as plane glass.

4. This is true statement. If we place the above combination in a medium of refractive index greater than this condition.

Question 12.
A lens of particular focal length is made from a given glass by adjusting radius of curvature. The formula applied in this case is lens maker’s formula
1. Write down lens maker’s formula
2. Derive lens maker’s formula considering refraction at a spherical surface
3. Explain the following facts based on lens maker’s formula

• power of sun glasses is zero even though they are curved
• if a lens is immersed in water focal length increases

Answer:
1.

2. Refraction by a lens:
Lens Maker’s Formula (for a thin lens): Consider a thin lens of refractive index n₂ formed by the spherical surfaces ABC and ADC. Let the lens is kept in a medium of refractive index n₁. Let an object ‘O’ is placed in the medium of refractive index n₁. Hence the incident ray OM is in the medium of refractive index n₁ and the refracted ray MN is in the medium of refractive index n₂.

The spherical surface ABC (radius of curvature R,) forms the image at l₁. Let ‘u’ be the object distance and ‘v₁’ be the image distance.
Then we can write,

Adding eq (1) and eq (2) we get

Dividing throughout by n₁, we get

if the lens is kept in air, \(\frac{n_{2}}{n_{1}}\) = n
So the above equation can be written as,

From the definition of the lens, we can take, when u = ∞, f = v
Substituting these values in the eq (3), we get

This is lens maker’s formula
\(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) ____(5)
For convex lens.
f = +ve, R₁ = +ve, R₂ = – ve

For concave lens,
f = -ve, R₁ = -ve, R₂ = +ve

Lens formula
From eq(4),

From eq(5)

From these two equations, we get

Linear magnification :
If h_o is the height of the object and h_i is the height of the image, then linear magnification.

3.
a. R₁ = R, R₂ = +R

power of lens, P = 0

b. We know

The above equation shows when n₁ increases f decreases the refractive index of water is greater than air. Hence when we place a lens in water, focul length decreases.

Students can Download Chapter 10 Wave Optic Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 10 Wave Optic

Plus Two Physics Wave Optic NCERT Text Book Questions and Answers

Question 1.
What is the shape of the wavefront in each of the following cases:

1. Light diverging from a point source.
2. Light emerging out of a convex lens when a point source is placed at its focus.
3. The portion of the wavefront of light from a distant star intercepted by the earth.

Answer:

1. It is spherical wavefront,
2. It is plane wavefront.
3. Plane wavefront (a small area on the surface of a large sphere is nearly planar).

Question 2.
What is the Brewster angle for to glass transition? (refractive index glass = 1.5).
Answer:
Given µ = 1.5, θ = ?
Since µ = tan θ
∴ tan θ = 1.5
or θ = tan^-1 1.5 = 56.3°.

Question 3.
Light of wavelength 5000 Å falls on a reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?
Answer:
Given λ = 5000 Å = 5 × 10^-7m
The wavelength and frequency of reflected light remains same.
∴ Wavelength of reflected light,
λ = 5000 Å.
Frequency of reflected light,

= 6 × 10¹⁴HZ
The reflected ray is normal to incident if angle of incidence i = 45°.

Question 4.
Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.
Answer:
Given a = 4mm = 4 × 10^-3m.
λ = 400 nm = 4 × 10^-7m
Z_F = ?
∴ The minimum distance a beam of light has to travel before its deviation from straight line path becomes significant is called Fresnel distance Z_F
∴ Z_F = \(\frac{a^{2}}{\lambda}=\frac{16 \times 10^{-6}}{4 \times 10^{-7}}\) = 40 m.

Plus Two Physics Wave Optic One Mark Questions and Answers

Question 1.
The reddish appearance of the sun at sunrise and sunset is due to
(a) The scattering of light
(b) The polarisation of light
(c) The colour of the sun
(d) The colour of the sky
Answer:
(a) The scattering of light
Explanation: The reddish appearance of the sun at sunrise and sunset is due to the scattering of light.

Question 2.
Angle between the plane of vibration and plane of polarization is
(a) 30°
(b) 90°
(c) 60°
(d) 70°
Answer:
(b) 90°
Explanation: Angle between the plane of vibration and plane of polarization is 90°.

Question 3.
If yellow light emitted by sodium lamp in Young’s double-slit experiment is replaced by monochromatic blue of light of the same intensity
(a) fringe width will decrease
(b) fringe width will increase
(c) fringe width will remain unchanged
(d) fringes will becomes less intense
Answer:
(a) fringe width will decrease
Explanation: As β = \(\frac{\lambda D}{d}\) and λ_b < λ_γ
∴ Fringe width p will decrease.

Question 4.
A Young’s double-slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen
(a) straight line
(b) parabola
(c) hyperbola
(d) circle
Answer:
(a) straight line
Explanation: Straight line fringes are formed on screen.

Question 5.
Find the odd one and justify interference, diffraction, polarisation.
Answer:
Polarisation, because polarisation is possible only for transverse wave. So all other phenomenon are due to super position of waves.

Question 6.
State Malus law related to the intensity of light transmitted through the analyzer.
Answer:
The Malus law states that the intensity of the polarized light transmitted varies as the square of the cosine of the angle between the plane of transmission of the analyser and plane of polarizer.

Question 7.
What is the minimum thickness of a soap film needed for constructive interference in reflected light, if the light incident on the film is 750 nm? Assume that the refractive index for the film is µ = 1.33.
(a) 282 nm
(b) 70.5 nm
(c) 141 nm
(d) 387 nm
Answer:
(c) 141 nm
Explanation: Here, 2µt = \(\frac{\lambda}{2}\)

Question 8.
Young’s experiment is performed with light of wavelength 6000 Å wherein 16 fringes occupy a certain region on the screen’. If 24 fringes occupy the same region with another light, of wavelength λ, then λ is
(a) 6000 Å
(b) 4500 Å
(c) 5000 Å
(d) 4000 Å
Answer:
(d) 4000 Å
Explanation: n₁λ₁ = n₂λ₂

Plus Two Physics Wave Optic Two Mark Questions and Answers

Question 1.
Name the following wavefronts according to its nature.

1. Wave front due to point source.
2. Wave front due to fluorescent lamp
3. Emergent wavefront from a concave lens.
4. Emergent wavefront from a prism, when plane is incident on other face.

Answer:

1. Spherical wave front
2. Cylindrical wavefront
3. Diverging wavefront
4. Plane wavefront

Question 2.
Two coherent sources have intensities in the ratio 25: 16. Find the ratio of intensities of maxima to minima after the superposition of waves from the two source.
Answer:
I₁ = a₁² = 25, a₁ = 5
I₂ = a₂² = 16
a₂ = 4
Maximum intensity I_max = (a₁ + a₂)² = (5 + 4)² = 81
Minimum intensity I_min = (5 – 4)² = 1

Plus Two Physics Wave Optic Three Mark Questions and Answers

Question 1.
Fill in the blanks in three columns.

Answer:
(i) θ = 42°
(ii) P = 57°
(iii) Straight line
(iv) Planks constant
(v) µ_r = 1.2
(vi) Paramagnetic.

Question 2.
Match the following suitably.

Answer:

Question 3.
Match the following

A	B
1. Colour of sky	1.  Interference
2. Rainbow	2. scattering
3. different colours seen in soap bubbles	3. Dispersion
	4. Diffraction
	5. Coherence
	6. Looming

Answer:

A	B
1. Colour of sky 2. Rainbow 3. different colours seen in soap bubbles	2. scattering 3. Dispersion 1. Interference

Question 4.
A plane wave front is entering a lens is given in the figure

1. What is meant by a wave front
2. What are different types of wave fronts
3. Complete the diagram and draw the refracted wave front.

Answer:

1. Locus of all points having same phase of vibration is called wavefront.
2. Spherical wavefront, cylindrical wavefront plane wave front
3. Wave frond through a thin convex lens:

Consider a plane wave passing through a thin convex lens. The central part of the incident plane wave travels through the thickest portion of lens. Hence central part get delayed. As a result the emerging wavefrond has a depression at the centre. Therefore the wave front becomes spherical and converges to a point F.

Plus Two Physics Wave Optic Four Mark Questions and Answers

Question 1.
Thomas young successfully conducted double-slit experiment and explained the interference phenomenon using Huygens principle.

1. State Huygens wave theory.
2. State Huygens principle arrive at Snell’s law of refraction.
3. In the word of Hyugens “Light propagates as longitudinal waves” comment on the above statement.

Answer:
1. According to Huygen’s principle

• Every point in a wavefront act as a source of secondary wavelets.
• The secondary wavelets travel with the same velocity as the original value.
• The envelope of all these secondary wavelets gives a new wavefront.

2. Huygen’s principle: According to Huygen’s principle:

• Every point in a wavefront acts as a source of secondary wavelets.
• The secondary wavelets travel with the same velocity as the original value.
• The envelope of all these secondary wavelets gives a new wavefront.

Refraction of a plane wave. (To prove Snell’s law):
AB is the incident wavefront and c₁ is the velocity of the wavefront in the first medium. CD is the refracted wavefront and c₂ is the velocity of the wavefront in the second medium. AC is a plane separating the two media.

The time taken for the ray to travel from Pto R is

O is an arbitrary point. Hence AO is a variable. But the time to travel a wavefront from AB to CD is constant. In order to satisfy this condition, the term containing AO in eq.(2) should be zero.

where ¹n₂ is the refractive index of the second medium w.r.t. the first. This is the law of refraction.

3. Light wave cannot be longitudinal as it exhibit polarisation.

Question 2.
A long narrow slit is illuminated by blue light and the diffraction pattern is obtained on a white screen.

1. How the width of bands change as the distance from the centre increases?
2. What happens to the width of pattern, if yellow light is used instead of blue light?
3. In a double-slit experiment, the slits are separated by 0.03cm and the screen is placed 1.5m away. The distance between the central fringe and the fourth bright fringe is 1 cm. Determine the wavelength of light used in the experiment.

Answer:
1. Width of bands decreases from the centre of the bands

2. The pattern expands or bandwidth increases

3. x_n = \(\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\)
d = 0.03cm = 0.03 × 10^-2 m
D = 1.5m, n = 4, x_n = 1cm = 1 × 10^-2m

Question 3.
Light of wavelength 589 nm is used to view an object under a microscope. The aperture of the objective has a diameter of 0.900cm.

1. What do you mean by limit of resolution of an optical instrument?
2. What is the limiting of resolution of the above microscope?
3. What happens to the limit of resolution if the objective is immersed in oil? Explain.

Answer:
1. The minimum distance between two objects at which they can be observed as separated by an optical instrument is called the limit of resolution of the instrument.

2. Limit of resolution ∆θ = \(\frac{1.22 \lambda}{D}=\frac{1.22 \times 589 \times 10^{-9}}{0.9 \times 10^{-2}}\)
= 7.98 × 10^-5 rad.

3. Decreases. Refractive index of oil is more than that of air. Hence wavelength of the light decreases.

Question 4.
The law of refraction \(\frac{\sin i}{\sin r}=\frac{V_{1}}{V_{2}}\)

1. This law is called
2. Prove this law based on Huygiens wave theory.

Answer:
1. Snells law

2. Refraction of a plane wave. (To prove Snell’s law):
AB is the incident wavefront and c₁ is the velocity of the wavefront in the first medium. CD is the refracted wavefront and c₂ is the velocity of the wavefront in the second medium. AC is a plane separating the two media.

The time taken for the ray to travel from Pto R is

O is an arbitrary point. Hence AO is a variable. But the time to travel a wavefront from AB to CD is constant. In order to satisfy this condition, the term containing AO in eq.(2) should be zero.

where ¹n₂ is the refractive index of the second medium w.r.t. the first. This is the law of refraction.

Question 5.
A long narrow slit is illuminated by blue light and the diffraction pattern is obtained on a white screen.

1. How the width of bands change as the distance from the centre increases? (1)
2. What happens to the width of pattern, if yellow light is used instead of blue light? (1)
3. In a double-slit experiment, the slits are separated by 0.03cm and the screen is placed 1.5m away. The distance between the central fringe and the fourth bright fringe is 1 cm. Determine the wavelength of light used in the experiment. (2)

Answer:
1. Width of bands decreases from the centre of the bands

2. The pattern expands or bandwidth increases

3. x_n = \(\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\)
d = 0.03cm = 0.03 × 10^-2 m
D = 1.5m, n = 4, x_n = 1cm = 1 × 10^-2m

Question 6.
Light of wavelength 589 nm is used to view an object under a microscope. The aperture of the objective has a diameter of 0.900cm.

1. What do you mean by limit of resolution of an optical instrument? (1)
2. What is the limiting of resolution of the above microscope? (1)
3. What happens to the limit of resolution if the objective is immersed in oil? Explain. (2)

Answer:
1. The minimum distance between two objects at which they can be observed as separated by an optical instrument is called the limit of resolution of the instrument.

2. Limit of resolution

= 7.98 × 10^-5rad.

3. Decreases. Refractive index of oil is more than that of air. Hence wavelength of the light decreases.

Question 7.
A beam of light, with intensity I₀, is passing through a polarizer and an analyzer as shown in figure.

1. State Malus law related to the intensity of light transmitted through the analyzer. (1)

2. If 6 = 45°, what is the relation of the intensities of original light and transmitted light after passing through the analyzer? (2)

3. Which of the following waves can be polarized

• X-rays
• sound waves. Why? (1)

Answer:
1. The Malus law states that the intensity of the polarized light transmitted varies as the square of the cosine of the angle between the plane of transmission of the analyser and plane of polarizer.

2.

3. X-rays. Because it is a transverse wave.

Plus Two Physics Wave Optic Five Mark Questions and Answers

Question 1.
Consider a point source emitting waves uniformly in all directions.

1. Draw two wave fronts very near to the point source. (1)
2. Using Huygen’s principle, prove that angle of incidence is equal to angle of reflection. (3)
3. What is the shape of a plane wave front after passing through a thin convex lens? (1)

Answer:
1.

2.

AB is the incident wavefront and CD is the reflected wavefront, ‘i’ is the angle of incidence and r’ is the angle of reflection. Let c₁ be the velocity of light in the medium. Let PO be the incident ray and OQ be the reflected ray.
The time taken for the ray to travel from P to Q is

O is an arbitrary point. Hence AO is a variable. But the time to travel for a wave front from AB to CD is a constant. So eq.(2) should be independent of AO. i.e., the term containing AO in eq.(2) should be zero.
∴ \(\frac{A O}{C_{1}}\)(sin i – sin r) = 0
sin i – sin r = 0
sin i – sin r
i = r
This is the law of reflection.

3. Spherical wavefront.

Question 2.
A slit S is illuminated by a monochromatic suorce of light to give two coherent sources P₁ and P₂. These give a dark band at the point R on the screen as shown in figure.

1. Write the formula to find out bandwidth. (1)
2. What relationship must exist between the length P₁R and P₂R? (1)
3. Can interference fringes be produced by using two identical bulbs? (1)
4. If the distance between P₁ and P₂ is 1 mm and the screen is placed in 1m away, what is the fringe separation for a light of wavelength 5 × 10^-7m?(2)

Answer:
1. β = \(\frac{\lambda D}{d}\)

2. The condition for dark fringe is P₂R – P₁R = (2n -1)\(\frac{\lambda }{2}\) where n = 1,2, 3, etc.

3. No. They are not coherent sources.

4.

Question 3.
In interference, when light energy superimpose with a phase difference 180°, darkness occurs.

1. Two sources which can give sustained interference pattern is said to be …….
2. Does interference phenomenon violate law of conservation of energy? Justify.
3. Modify the expression for bandwidth in terms of refractive index of medium between slit and screen.

Answer:
1. Coherent sources.

2. No. In interference, only energy is redistributed.

3. We know refractive index n = \(\frac{c}{v}=\frac{\lambda}{\lambda_{1}}\)
λ₁ = \(\frac{\lambda}{n}\) substituting this in the expression for bandwidth \(\left(\beta=\frac{\lambda, \mathrm{D}}{\mathrm{d}}\right)\)
we get \(\beta=\frac{\lambda D}{n d}\).

Question 4.
Figure below shows a version of Young’s Experiment performed by directing a beam of electrons on a double slit. The screen reveals a pattern of bright and dark fringes similar to an interference pattern produced when a beam of light is used.

1. Which property of electron is revealed in this observation.
2. If the electrons are accelerated by a p.d. of 54v, what is the value of wavelength associated with electrons.
3. In similar experiment, if the electron beam is re-placed by bullets fired from a gun, no interference pattern is observed. Give reason.

Answer:
1. Dual nature or wave nature.

2.

3. λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\). Since the mass ofthe bullet is very much greater than the mass of electron, the de Broglie wavelength is not appreciable.

Question 5.
A. In young’s double-slit experiment two slits are illuminated by real monochromatic light source.

1. If one of the slits is closed, what will be the observation on the screen?
2. Arrive at an expression for bandwidth of interference fringes, when both the slits are open.
3. What happens to the bandwidths, if the experimental arrangement is immersed in water?

Answer:
1. Single slit diffraction pattern.

2. Expression for band width:

S₁ and S₂ are two coherent sources having wave length λ. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O’ is a point on the screen equidistant from S₁ and S₂.

Hence the path difference, S₁O – S₂O = 0
So at ‘O’ maximum brightness is obtained.
Let ‘P’ be the position of n^th bright band at a distance x_n from O. Draw S₁A and S₂B as shown in figure.
From the right angle ∆S₁AP
we get, S₁P² = S₁A² + AP²
S₁P² = D² + (X_n – d/2)²
= D² + X_n² – X_nd + \(\frac{\mathrm{d}^{4}}{4}\)
Similarly from ∆S₂BP weget,

S₂P² = S₂B² + BP²
S₂P² = D² + (X_n + d/2)²

S₂P² – S₁P² =2x_nd (S₂P + S₁P)(S₂P – S₁P) = 2x_nd
But S₁P ≈ S₂P ≈ D
∴ 2D(S₂P – S₁P) = 2x_nd
i.e., path difference S₂P – S₁P = \(\frac{x_{n} d}{D}\) ____(1)
But we know constructive interference takes place at P, So we can take
(S₂P – S₁P) = nλ
Hence eq(1) can be written as

Let x_n+1 be the distance of (n+1)th bright band from centre o, then we can write

∴ band width, b

β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
This is the width of the bright band. It is the same for the dark band also.

3. Since wavelength decreases, the bandwidth decreases.

Question 6.
Green light is incident at the Polarizing angle on a certain glass plate as shown in figure.

1. What do you mean by polarizing angle?
2. Indicate the polarization components on the reflected and refracted rays, by arrows and dots.
3. Find the refractive index of glass.

Answer:
1. The angle of incidence, at which incident light on a transparent medium, become completely plane polarized is known as polarizing angle.

2.

3. Polarizing angle = 90° – 32° = 58°, Refractive Index, n = tan 58° = 1.6.

Question 7.
Allow to lights rays to incident on a screen after passing through two slits.

1. Why light is passed through two slits.
2. Find the expression for fringe width
3. What happens to the pattern on the screen when the whole apparatus is dipped in water

Answer:
1. Because light has wave nature. Two slits gives effect of coherent sources.

2. Expression for band width:

S₁ and S₂ are two coherent sources having wave length λ. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O’ is a point on the screen equidistant from S₁ and S₂.

Hence the path difference, S₁O – S₂O = 0
So at ‘O’ maximum brightness is obtained.
Let ‘P’ be the position of n^th bright band at a distance x_n from O. Draw S₁A and S₂B as shown in figure.
From the right angle ∆S₁AP
we get, S₁P² = S₁A² + AP²
S₁P² = D² + (X_n – d/2)²
= D² + X_n² – X_nd + \(\frac{\mathrm{d}^{4}}{4}\)
Similarly from ∆S₂BP weget,
S₂P² = S₂B² + BP²
S₂P² = D² + (X_n + d/2)²

S₂P² – S₁P² =2x_nd (S₂P + S₁P)(S₂P – S₁P) = 2x_nd
But S₁P ≈ S₂P ≈ D
∴ 2D(S₂P – S₁P) = 2x_nd
i.e., path difference S₂P – S₁P = \(\frac{x_{n} d}{D}\) ____(1)
But we know constructive interference takes place at P, So we can take
(S₂P – S₁P) = nλ
Hence eq(1) can be written as

Let x_n+1 be the distance of (n+1)th bright band from centre o, then we can write

∴ band width, b

β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
This is the width of the bright band. It is the same for the dark band also.

3. The pattern shrinks (Band width decreases) if whole apparatus is dipped in water. (Because of high refractive index, velocity of light decreases. The wavelength also decreases and hence fringe width also reduce).

Question 8.
The double slit in Young’s experiment is replaced by a single narrow slit (illuminated by a monochromatic source). Then,

1. How the pattern of bands on the screen differ from the pattern due to double slit? (1)
2. Derive an expression for the bandwidth of the central fringe. (3)
3. Draw a graph which shows the variation of intensity of light with distance. (1)

Answer:
1. A broad pattern with a central bright region is seen. On both sides, there are alternate dark and bright regions of decreasing intensity.

2. Consider a point P, on the screen at which wavelets travelling in a direction making an angle θ with CO are brought to focus by the lens are shown in figure.

These wavelets travel unequal distances in reaching the point P₁. Hence these waves are not in phase. The wavelets from the points A and B reaching P₁ are having a path difference, BP₁ – AP₁ = BN = a sinθ.

This path difference equals to λ. Then for each point in the upper half AC of the slit, there is a corresponding point in the lower half CB such that the wavelets from these two points reach at P₁ with a path difference of λ/2.

These wavelets interfere destructively to make the intensity at P₁ minimum. The point P₁corresponds to first minima. Condition for first minima is a sinθ = λ.

Thus in general, the minima will occur when the path difference = asinθ = λ where n = 1,2,3. Thus minima are formed on both sides of O, i.e. the central maxima. In between minima, other maxima called secondary maxima are formed. Secondary maxima will be at those points for which the path difference for the rays is asinθ = (2n + 1)\(\frac{\lambda}{2}\).

The width of the central maximum is defined as the distance between the first minima on either side of the central maximum. For the first minimum, a sinθ = λ when θ is small sinθ = θ. i.e.a θ = λ.

3.

Question 9.
A slit S is illuminated by a monochromatic suorce of light to give two coherent sources P₁ and P₂. These give a dark band at the point R on the screen as shown in figure.

1. Write the formula to find out bandwidth, (1)
2. What relationship must exist between the length P₁R and P₂₁R? (1)
3. Can interference fringes be produced by using two identical bulbs? (1)
4. If the distance between P₁ and P₂ is 1 mm and the screen is placed in 1m away, what is the fringe separation for a light of wavelength 5 × 10^-7m?(2)

Answer:
1. β = \(\frac{\lambda D}{d}\)

2. The condition for dark fringe is P₂R – P₁R = (2n -1)\(\frac{\lambda }{2}\) where n = 1,2, 3, etc.

3. No. They are not coherent sources.

4.

Question 10.
The Photograph given below is obtained by passing a LASER beam on a pain of closely spaced slits.

1. Identify this pattern.
2. Obtain an expression for bandwidth of the pattern.
3. In the double slit experiment using wavelength 5461 A⁰, the fringe width measured is 0.15mm. By keeping the same arrangement, the fringe width is measured for an unknown wavelength is 0.12mm. Find the unknown wave length.
4. If you change the LASER light from red to blue, what will happen to the space between the pattern shown in photograph. Justify.

Answer:
1. Interference

2. Expression for band width:

S₁ and S₂ are two coherent sources having wave length λ. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O’ is a point on the screen equidistant from S₁ and S₂.

Hence the path difference, S₁O – S₂O = 0
So at ‘O’ maximum brightness is obtained.
Let ‘P’ be the position of n^th bright band at a distance x_n from O. Draw S₁A and S₂B as shown in figure.
From the right angle ∆S₁AP
we get, S₁P² = S₁A² + AP²
S₁P² = D² + (X_n – d/2)²
= D² + X_n² – X_nd + \(\frac{\mathrm{d}^{4}}{4}\)
Similarly from ∆S₂BP weget,
S₂P² = S₂B² + BP²
S₂P² = D² + (X_n + d/2)²

S₂P² – S₁P² =2x_nd (S₂P + S₁P)(S₂P – S₁P) = 2x_nd
But S₁P ≈ S₂P ≈ D
∴ 2D(S₂P – S₁P) = 2x_nd
i.e., path difference S₂P – S₁P = \(\frac{x_{n} d}{D}\) ____(1)
But we know constructive interference takes place at P, So we can take
(S₂P – S₁P) = nλ
Hence eq(1) can be written as

Let x_n+1 be the distance of (n+1)th bright band from centre o, then we can write

∴ band width, b

β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
This is the width of the bright band. It is the same for the dark band also.

3.

λ₁ = 5461 A° = 5461 × 10^-10 m
β₁ = 0.15mm = 0.15 × 10^-3m
β₂ = 0.12mm = 0.12 × 10^-3m.

4. Wave length of blue is less than that of red. Hence β decreases.

Question 11.
According to a principle, at a particular point in a medium, the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each of the waves.

1. Name of the principle.
2. Derive an expression for the bandwidth in Young’s double-slit experiment.

Answer:
1. Superposition principle.

2. Expression for band width:

S₁ and S₂ are two coherent sources having wave length λ. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O’ is a point on the screen equidistant from S₁ and S₂.

Hence the path difference, S₁O – S₂O = 0
So at ‘O’ maximum brightness is obtained.
Let ‘P’ be the position of n^th bright band at a distance x_n from O. Draw S₁A and S₂B as shown in figure.
From the right angle ∆S₁AP
we get, S₁P² = S₁A² + AP²
S₁P² = D² + (X_n – d/2)²
= D² + X_n² – X_nd + \(\frac{\mathrm{d}^{4}}{4}\)
Similarly from ∆S₂BP weget,
S₂P² = S₂B² + BP²
S₂P² = D² + (X_n + d/2)²

S₂P² – S₁P² =2x_nd (S₂P + S₁P)(S₂P – S₁P) = 2x_nd
But S₁P ≈ S₂P ≈ D
∴ 2D(S₂P – S₁P) = 2x_nd
i.e., path difference S₂P – S₁P = \(\frac{x_{n} d}{D}\) ………1)
But we know constructive interference takes place at P, So we can take
(S₂P – S₁P) = nλ
Hence eq(1) can be written as

Let x_n+1 be the distance of (n+1)th bright band from centre o, then we can write

∴ band width, b

β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
This is the width of the bright band. It is the same for the dark band also.