Class 10 Physics Chapter 2 Important Questions Kerala Syllabus Lenses

The comprehensive approach in SCERT Class 10 Physics Solutions Chapter 2 Lenses Important Questions with Answers ensure conceptual clarity.

SSLC Physics Chapter 2 Important Questions Kerala Syllabus

Lenses Class 10 Important Questions

Question 1.
The midpoint of the lens is ……………………
Answer:
Optic centre

Question 2.
Find the relation in the first pair and complete the second.
Images that can be captured on the screen: Real image
Images that cannot be captured on the screen: ……………………..
Answer:
Virtual image

Question 3.
Write the lens equation.
Answer:
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
f-focal length,v-image distance, u-object distance

Question 4.
The area of the lens through which light passes ……………………
Answer:
Aperture

Question 5.
Find the relation in the first pair and complete the second.
Focal length : metre
Power of lens : …………………
Answer:
Dioptre

Question 6.
Find the odd one out
(Optic centre,magnification,aperture,centres of curvature)
Answer:
Magnification

Question 7.
The lens that forms only virtual images is …………………….
(convex lens, concave lens)
Answer:
concave lens

Question 8.
Find the correct lens equation.
a) (1/u) – (1/v) = 1/f
b) (1/v) – (1/u) = 1/f
c) (1/v) + (1/u)= 1/f
d) (1/f) – (1/u) = (1/v)
Answer:
b) (1/v) – (1/u) = 1/f

Question 9.
If the image formed is virtual and erect its magnification is …………………….
Answer:
Positive

Question 10.
The image formed by a convex lens is inverted and diminished. Then the object must be placed: | At 2F, Beyond 2F, Between F and 2F, At F ]
Answer:
Beyond 2F

Question 11.
According to Cartesian Sign Conventions, sign of focal length of a concave lens is (Model 2023)
Answer:
Negative

Question 12.
What is the power of a lens having focal length 100 cm.
Answer:
Focal length of concave lens = -100 cm = -1m
Power of the lens p = 1/f = 1/-1 = -1 D
If the power is negative lens can be identified as concave.

Question 13.
Is it possible for a convex lens to form a virtual image? Explain with the help of a figure.
Answer:
It is possible.

Question 14.
State whether the following statements are correct or incorrect. Correct the incorrect statements
According to cartesian sign convention
a) When the magnification is positive, the image is inverted.
b) When the magnification is negative, the image is erect.
c) When the magnification is less than one , the image is diminished.
d) Negative sign of magnification indicates that the image is inverted and real.
Answer:
a) Incorrect statement
When the magnification is positive, the image is erect

b) Incorrect statement
When the magnification is negative, the image is inverted.

c) Correct statement

d) Correct statement

Question 15.
Draw the ray diagram showing the image formation in a convex lens when the object is placed between F and 2F.
Answer:

Question 16.
What do you mean by virtual image?
Answer:
Images that cannot be captured on a screen, but can only be seen are virtual images.

Question 17.
Draw a ray diagram showing the virtual image formation in a concave lens.
Answer:

Question 18.
Observe the figure.

a) Write the values of u and f by using Cartesian Sign Convention.
b) Calculate the distance of the image from the lens? (Model 2020)
Answer:
a) u = -45 cm f = -30 cm
b)

Question 19.
Analyse the figure. An object is placed between F and 2F of a convex lens.

a) Copy the diagram and complete to show the image formation.
b) Where must be the object be placed to get a real image of same size as that of the object (March 2022)
Answer:

b) At 2F

Question 20.
Complete the table given below

Position of object	Position of image	Characteristics of image
…………………	Between 2F and F	Diminished, inverted, real
At 2F	……………………	Same size, inverted, real
Between 2F and F	Beyond 2F	…………………..
…………………..	Infinity (Far away)	Magnified, inverted, real
Between F and lens	Between 2F and F on the same side of the object	………………….

Answer:

Position of object	Position of image	Characteristics of image
Beyond 2F	Between 2F and F	Diminished, inverted, real
At 2F	At 2F	Same size, inverted, real
Between 2F and F	Beyond 2F	Magnified, inverted, real
At F	Infinity (Far away)	Magnified, inverted, real
Between F and lens	Between 2F and F on the same side of the object	Magnified, erect, virtual

Question 21.
The focal length of a convex lens is 20 cm. An object of height 2 cm is placed on its optic axis is located 40 cm away from the lens. Calculate the distance from the lens to the image. What will be the magnification in this case? What are the characteristics of the image formed?
Answer:
f = 20 cm
h_o = 2 cm
The object is located at 2f (at 40 cm) of the convex lens. Here, the distance from the lens to the image v = 40 cm.
Magnification = -1
Characteristics of the image: same size, inverted, real

Question 22.
Select the appropriate ones from the box based on statements given below.

Telescope, Compound microscope, objective, eyepiece

a) A device that magnifies objects.
b) A device to see distant objects clearly.
c) The lens through which the image is observed
d) Lens placed near the object to be observed
Answer:
a) compound microscope
b) telescope
c) eyepiece
d) objective

Question 23.
Tabulate the given statements
• Thinner in the middle
• Thinner at the edges
• Thicker in the middle
• Thicker at the edges
• Shows the objects magnified
• Shows the object as diminished
Answer:

Convex lens	Concave lens
• Thicker in the middle • Thinner at the edges • Shows the objects magnified	• Thinner in the middle • Thicker at the edges • Shows the object as diminished

Question 24.
Draw the ray diagram showing the image formation in a convex lens when the object is placed beyond 2F. Write down the position and characteristics of image.

Answer:
Position of image : Between 2F and F on other side of the lens
Characteristics of image:Inverted, diminished and real.

Question 25.
If the magnification is positive and less than one, which is the type of lens. Write down the nature of image.
Answer:
Magnification is less than one for concave lens.
Position of image: between F and lens on the same side
Characteristics of image diminished, virtual and erect

Question 26.
If the magnification is greater positive and greater than one, which is the type of lens .Write down the nature of image.
Answer:
Magnification is greater than one for convex lens.
Position of image: on the same side of the object.
Characteristics of image: magnified, virtual and erect

Question 27.
Which are the cartesian sign convention rules?
Answer:

• All distances should be measured from the optic centre of the lens.
• Distances measured in the same direction as the incident ray should be considered positive and those in the opposite direction should be considered negative.
• Distances measured above the optic axis should be considered positive and those below should be considered negative.
• Cartesian sign convention can be used to solve mathematical problems using general equations in different contexts. There is no need to consider whether the object is on the left or right side of the lens.

Question 28.
The nature of images formed by two lenses are given
i) An erect and magnified virtual image
ii) An erect and diminished virtual image
a) What type of lens is used in each case?
b) On using which of these lenses will we get an image having the same size as the object? What is the position of the object?
Answer:
a) i) Convex lens
ii) Concave lens
b) Convex lens at 2F

Question 29.
An object of height 3 cm is placed in front of a convex lens of focal length 20 cm at a distance of 30 cm.
a) What is the distance to the image formed?
b) What is the nature of the image?
c) What is the height of the image?
Answer:
a) u = -30 cm
v = ?
f = +20 cm

b) Beyond 2F, Real, Inverted

c) m = \(\frac{v}{u}\)
v = 60 cm
u = -30 cm
m = \(\frac{60}{-30}\) = -2
h₀ = 3 cm
m = \(\frac{h_{\mathrm{i}}}{h_0}\)
-2 = \(\frac{h_{\mathrm{i}}}{3}\)
h_i = -6 cm
Image is inverted.

Question 30.
a) What is power of a lens?
b) In the prescription of a doctor, the power of lens is noted as +2 D. Find the focal length.
Answer:
a) Reciprocal of focal length expressed in metre
b) P = \(\frac{1}{f}\)
∴ f = \(\frac{1}{P}\) = \(\frac{1}{+2}\) = 0.5 m = 50 cm

Question 31.
a) Write lens equation.
b) When an object is placed at a distance of 10 cm from a concave lens its image is formed at a distance of 6 cm from the lens. Calculate the focal length of the lens?
Answer:
a) \(\frac{1}{f}\) = \(\frac{1}{v}\) – \(\frac{1}{u}\)

b) f = \(\frac{u v}{u-v}\)
u = -10 cm
v = -6 cm
= \(\frac{-10 \times-6}{-10-6}\)
= \(\frac{+60}{-4}\) = -15 cm

Question 32.
Focal length of a concave lens is 20 cm . If the distance to the image is 12 cm, find out the distance to the object from the lens.
Answer:
f = -20 cm
v = -12 cm
u = ?

Question 33.
Define the following terms of a lens.
a) Optic centre
b) Centre of curvature
c) Principal focus of convex lens
d) Focal length of concave lens
Answer:
a) The midpoint of a lens

b) The centre of the imaginary spheres of which the sides of the lens are parts.

c) Light rays incident parallel and close to the optic axis after refraction converges to a point on the optic axis of a convex lens. This is the principal focus of a convex lens.

d) The distance from the principal focus of a concave lens to its optic centre.

Question 34.
a) If the image obtained from a convex lens is erect and enlarged:
i. Image is formed at: [Same side of the object / Opposite side of the object]
ii. Write any one application of this type of image formation.
b) Write any two characteristics of the image formed by a convex lens when the object is placed at the following positions. (March 2021)
i. At infinity
ii. Between 2F and F
Answer:
a) i. Same side of the object
ii. Used in simple microscope, telescope, compound microscope, reading lens etc.

b) i)
Nature of Image: Highly diminished, real and inverted
ii)
Nature of image: Enlarged, real and inverted

Question 35.
An object is placed at a distance of 15 cm from a lens of focal length + 10 cm.
a) Which type of lens is used here?
b) Calculate the image distance.
c) What is the nature of the image formed? (March 2023)
Answer:
a) Convex lens

b) u = -15 cm f = + 10 cm

c) Real and inverted

Question 36.
Find the relation between the first word pair and complete the second pair.
Mirror: Reflection
Lens : ………………….
Answer:
Refraction

Question 37.
Observe the following figure and answer the following question.

a) Identify the lens given in the figure.
b) What is the nature of the principal focus of this lens.
Answer:
a) Concave lens
b) Virtual

Question 38.
Why does a concave lens have a virtual focus?
Answer:
Concave lens cannot converge the incident rays at a point. A real focus is only formed where light rays converge. There fore it has a virtual focus.

Question 39.
Complete the sentences
a) Convex lens has a ………………………. focus
b) A convex lens has ………………………. principal focii(1, 2)
Answer:
a) real
b) 2

Question 40.
Choose the statements related to concave and convex lens.
i. The light ray diverges
ii. The light ray converges
iii. When the letters are viewed through this lens, by moving the lens to one side, the letters appear to move in the opposite direction.
iv. When the letters are viewed through this lens, by moving the lens to one side the letters appear to move in the same direction.
Answer:
Concave lens – i, iv
Convex lens – ii, iii

Question 41.
Complete the table for convex lens.

Position of object	Position of image
2F	…………(a)……………
Between F and lens	…………..(b)………….

Answer:
(a) 2F
(b) At same side of the object.

Question 42.
You need to project image of an object on a wall. Which lens would you choose? Why?
Answer:
Convex lens is to be choosed. It forms real images which can be captured on a screen.

Question 43.
Correct the following figures if wrong

Answer:

Question 44.
Nature of some images are given below. Find out that suits for the real and virtual images and then write it in separate column.
a) Inverted image
b) not obtained on a screen
c) Obtained on a screen
d) The light rays really intersect to each other.
e) Erect image
f) The image distance cannot directly measured.
Answer:
Real images: a, c, d
Virtual image : b, e, f

Question 45.
When an object is placed at a distance of 15 cm from a convex lens ,a real image is formed at a distance of 30 cm. What is the focal length of the lens.
Answer:
u = -15 cm,v = +30 cm
f = \(\frac{u v}{u-v}\)
= \(\frac{-15 \times+30}{-15-+30}\) = \(\frac{-15 \times 30}{-45}\) = + 10 cm

Question 46.
The magnification in the case of a lens is -ve
a) The image formed will be ………………. (erect/inverted)
b) The lens must be ……………….. (concave/convex)
Answer:
a) inverted
b) convex lens

Question 47.
Write the measures given in the figures by new Cartesian sign convention.

a) Object distance (u)
b) Image distance (v)
c) Height of the object (OB)
d) Height of the image (IM)
e) What is the focal length of this lens
f) Find magnification of image.
Answer:
a) u = -25 cm
b) v = +100 cm
c) OB = +1 cm
d) IM = -4 cm
e) u = -25 cm
v = +100 cm
f = \(\frac{u v}{u-v}\) = \(\frac{-25 \times 100}{-25-100}\) = \(\frac{-2500}{-125}\) = +20 cm

f) m = v/u
m = 100/-25 = -4

Question 48.
Calculate the height of image if magnification is -3 and height of object is 2 cm
Answer:
h₀ = 2 cm, m = h_i / h₀ = -3
h_i = -3 × 2 = -6 cm

Question 49.
Calculate the power of a lens of focal length +15 cm.
Answer:
f = +15 cm = 0.15m
P = 1/f = 1/0.15 = 6.66 D

Question 50.
State whether the given statements are correct or incorrect and correct the incorrect statement.
a) The unit of power is dioptre (D)
b) Power of a lens is the reciprocal of focal length expressed in centimetres
Answer:
a) Correct statement
b) Incorrect statement
Corrected statement – Power of a lens is the reciprocal of focal length expressed in metres

Question 51.
When a person suffering from defective vision met a doctor, in the prescription it was written as +3 D for buying spectacles. Which is the lens to be used in spectacles? How did you identify it.?
Answer:
Convex lens is to be used. The positive sign of power indicates that the lens is convex.

Question 52.
The power of a concave lens is ……………………. (positive, negative)
Answer:
negative

Question 53.
State whether the given statements are correct or incorrect and correct the incorrect statement.
a) Objective in a compound microscope is a concave lens
b) Eye piece of compound microscope forms magnified, virtual and erect image
Answer:
a) Incorrect statement
Corrected statement: Objective in a compound microscope is a convex lens
b) Correct statement

Question 54.
Complete the following sentences
a) Focal length of the objective in a refracting telescope is
(lesser / greater)
b) In a telescope, the objective forms a ……………………… image of a distant object
(small, real and inverted; magnified, real, inverted)
Answer:
a) greater
b) small, real and inverted

Question 55.
The functioning of a telescope is based on …………………….. of light
(Refraction, reflection)
Answer:
Refraction

Question 56.
What is the difference between a refracting telescope and a compound microscope? How does magnification work in both devices?
Answer:
A refracting telescope is used to view distant objects (like stars), while a compound microscope is used to view very small objects (like cells). Both devices use two lenses: one to form the image (objective lens) and another to magnify it (eyepiece lens). However, the telescope magnifies distant objects. while the microscope magnifies tiny, close objects.