74 MCQs on Light – Reflection and Refraction Chapter 10 | Class 10 Science MCQs| Important for SEBA / CBSE
In this article, you will get high-quality MCQs with answers and suitable explanations on the Light – Reflection and Refraction Chapter 10 Class 10 Science. It includes
- “SEBA 10 Years PYQ MCQs on Light – Reflection and Refraction“
- “Assertion-Reason MCQs on Light – Reflection and Refraction “
- “Statement-Based MCQs on Light – Reflection and Refraction“
“Match the Following type MCQs on Light – Reflection and Refraction“
To score 100 % marks in your HSLC/CBSE Board exam, we have provided 74 MCQs on Light – Reflection and Refraction in this article. We also provided simple Explanations so you can understand the chapter easily.
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74 MCQs on Light – Reflection and Refraction Chapter 10
1.If a virtual, erect and enlarged image is formed by a lens, then which of the following statement is correct? [HSLC 2026]
(A) It is a concave lens and the object is placed between optical centre and focus of the lens.
(B) It is a convex lens and the object is placed between focus and the centre of curvature of the lens.
(C) It is a concave lens and the object is placed between focus and the centre of curvature of the lens.
(D) It is a convex lens and the object is placed between optical centre and the focus of the lens.
View Answer...
Explanation:
A convex lens forms a virtual, erect and enlarged image only when the object is placed between the optical centre and the principal focus of the lens.
A concave lens always forms a virtual and erect image, but it is diminished, not enlarged.
2. An object of size 4 cm height is placed in between centre of curvature and focus of a concave mirror of focal length 10 cm. The nature and size of the image will be – [HSLC 2026]
(A) real, inverted and bigger than 4 cm.
(B) real, inverted and smaller than 4 cm.
(C) virtual, erect and bigger than 4 cm.
(D) virtual, erect and smaller than 4 cm.
View Answer...
Explanation:
When an object is placed between the focus ($F$) and the centre of curvature ($C$) of a concave mirror, the image formed is always real, inverted, and enlarged. Since the original height is 4 cm, the image size will be bigger than 4 cm.
3. A student conducts an activity using a concave mirror with focal length of 10 cm. He places an object 15 cm from the mirror. Where is the image likely to form? [HSLC 2025]
(A) At 6 cm behind the mirror
(B) At 30 cm behind the mirror
(C) At 6 cm in front of the mirror
(D) At 30 cm in front of the mirror
View Answer...
Explanation:
We know that : $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
For a concave mirror, focal length ($f$) = $-10\text{ cm}$ and object distance ($u$) = $-15\text{ cm}$
so: $\frac{1}{-10} = \frac{1}{v} + \frac{1}{-15}$
$\frac{1}{v} = \frac{1}{15} – \frac{1}{10} = \frac{2 – 3}{30} = -\frac{1}{30}$
So, $v = -30\text{ cm}$.
Negative sign means the image is formed in front of the mirror.
So, the image is formed 30 cm in front of the mirror.
4. A concave lens has a focal length of 20 cm. What is the power of the lens? [HSLC 2025]
(A) −5 dioptre
(B) −0.05 dioptre
(C) +0.05 dioptre
(D) +5 dioptre
View Answer...
Explanation:
Power of a lens is calculated as $P = \frac{1}{f\text{ (in meters)}}$.
Given, focal length $f= 20\text{ cm} = 0.2\text{ m}$.
Here, for a concave lens, focal length is negative: so $f = -0.2\text{ m}$.
So, Power $P = \frac{1}{-0.2} = -5\text{ D}$.
5. The refractive indices of four mediums P, Q, R, S are 1.23, 1.5, 1.25 and 2 respectively. Through which medium, the speed of light is maximum? [HSLC 2025]
(A) P
(B) Q
(C) R
(D) S
View Answer...
Explanation:
We know that, smaller refractive index ⇒ greater speed of light.
Given:
- P = 1.23
- Q = 1.5
- R = 1.25
- S = 2
The smallest refractive index is 1.23, so light travels fastest in P.
6. A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be- [HSLC 2024]
(A) both convex
(B) both concave
(C) the mirror is concave and the lens is convex
(D) the mirror is convex and the lens is concave
View Answer...
Explanation:
By sign convention:
- For a convex mirror, focal length is positive
- For a concave mirror, focal length is negative
- For a convex lens, focal length is positive
- For a concave lens, focal length is negative
Since both have focal length −15 cm:
- mirror → concave
- lens → concave
7. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object? [HSLC 2021, 24]
(A) Between the principal focus and the centre of curvature
(B) At the centre of curvature
(C) Beyond the centre of curvature
(D) Between the pole of the mirror and its principal focus
View Answer...
Explanation:
A concave mirror forms a virtual, erect and enlarged image only when the object is placed between the pole (P) and focus (F).
8. The SI unit of power of a lens is— [HSLC 2024]
(A) joule
(B) dioptre
(C) ohm
(D) ampere
View Answer...
9. Image formed by a concave mirror is – [HSLC 2023]
(A) Real and diminished
(B) Real and enlarged
(C) Virtual and enlarged
(D) Virtual and diminished
View Answer...
10. If the refractive index of glass is 1.5, then speed of light in glass is – [HSLC 2023]
(A) 2 × 10⁸ m/s
(B) 3 × 10⁵ m/s
(C) 2.25 × 10⁸ m/s
(D) 3 × 10⁸ m/s
View Answer...
Explanation:
We know that, the refractive index is $n = \frac{c}{v}$
- So, $v = \frac{c}{n}$
$v = \frac{3 \times 10^8}{1.5} = 2 \times 10^8\text{ m/s}$.
11. Where should an object be placed in front of a convex lens to get a real image of the same size as the object? [HSLC 2019, 22]
(A) At infinity
(B) At the principal focus
(C) At twice the focal length
(D) Between optical centre and principal focus
View Answer...
Explanation:
When we place an object exactly at twice the focal length ($2F_1$) of a convex lens, the lens forms an image on the other side at $2F_2$. This image is real, inverted, and exactly the same size as the object.
12. A student conducts an experiment convex lens. He places the object at a distance of 60 cm in front of the lens and observe that the image is formed at a distance of 30 cm behind the lens. What is the power of the lens?
(A) 0.005 dioptre
(B) 0.05 dioptre
(C) 5 dioptre
(D) 50 dioptre
View Answer...
Explanation:
We know that, $\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$.
Given, ($u$) = $-60\text{ cm}$
($v$) = $+30\text{ cm}$
$\frac{1}{f} = \frac{1}{30} – \frac{1}{-60}$
$ \frac{1}{f} = \frac{1}{30} + \frac{1}{60}$
$ \frac{1}{f} = \frac{2 + 1}{60} = \frac{3}{60} = \frac{1}{20}$
$f = 20\text{ cm}$. In meters, this is $0.2\text{ m}$.
So, Power $P = \frac{1}{f\text{ (in meters)}}$
$P = \frac{1}{0.2} = 5\text{ dioptre}$.
So the power is 5 dioptre.
13. Assertion (A): The refractive index of diamond is √6 and refractive index of liquid is √3. If the light travels from diamond to the liquid, it will initially reflected when the angle of incidence is 30°.
Reason (R): μ = 1/sin C (w.r.t. liquid), where μ is the refractive index of diamond
(A) A and R are true and R is the correct explanation of A
(B) A and R are true and R is not correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
View Answer...
Explanation: Let’s find the critical angle ($C$) for light moving from diamond to liquid:
Refractive index of diamond relative to liquid is $\mu = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}$.
Now, $ \mu= \frac{1}{\sin C }$
$\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$
which means $C = 45^\circ$.
For total internal reflection, angle of incidence must be greater than critical angle.
But here angle of incidence is 30°, which is less than 45°.
So total internal reflection will not occur.
Therefore,
Assertion is false
Reason is true
14. Magnification of plane mirror is +1. It means –
i) Image size = object size
ii) Image is virtual and erect
iii) Image laterally inverted
iv) Image distance = object distance from mirror
(A) i
(B) i, iv
(C) i, ii, iv
(D) i, ii, iii, iv
Explanation:
For a plane mirror, magnification is +1. This means:
(i) Image size = object size ✔
(ii) Image is virtual and erect ✔
(iii) Image is laterally inverted ✔
(iv) Image distance = object distance from mirror ✔
So all four statements are correct.
15. During refraction of light through a rectangular glass slab –
(i) Emergent ray is parallel to incident ray.
(ii) Angle of emergence equals angle of incidence.
(iii) Emergent ray is laterally shifted.
(iv) Angle of emergence is greater than angle of incidence.
(A) (i), (ii)
(B) (i), (iii), (iv)
(C) (i), (ii), (iii)
(D) (i), (iv)
Explanation:
(i) Emergent ray is parallel to incident ray. ✔
(ii) Angle of emergence equals angle of incidence. ✔
(iii) Emergent ray is laterally shifted. ✔
(iv) Angle of emergence is greater than angle of incidence. ✘
The correct set should be (i), (ii), (iii).
16. If a light ray moving in a straight line changes its direction by a definite acute angle, the phenomenon is called—
(i) Reflection
(ii) Refraction
(iii) Dispersion
(iv) Diffraction
(A) (i) and (iii)
(B) (i) and (ii)
(C) (i) only
(D) (ii) only
Explanation: Both reflection and refraction cause a straight light ray to change its direction by a definite acute angle.
17. Match the following and choose appropriate sequence of two column for a concave mirror:
| Position of object | Position of image |
| (i) At infinite | (A) At focus |
| (ii) Focus | (B) Centre of curvature |
| (iii) Centre of curvature | (C) At infinity |
| (iv) Between focus and pole | (D) Behind mirror |
(A) A-(ii), B-(iii), C-(i), D-(iv)
(B) A-(i), B-(iii), C-(ii), D-(iv)
(C) A-(iv), B-(iii), C-(ii), D-(i)
(D) A-(iv), B-(i), C-(iii), D-(ii)
Explanation:
Object position → image position for concave mirror:
At infinity → At focus
At focus → At infinity
At centre of curvature → Centre of curvature
Between focus and pole → Behind mirr
18. If the image formed by a mirror for all positions of the object placed in front of it is always diminished, erect and virtual, the mirror is:
(A) Both concave and convex
(B) Plane
(C) Concave
(D) Convex
View Answer...
Explanation:
A convex mirror always forms an image that is virtual, erect, and diminished than the object. This is why they are perfect for rear-view mirrors in cars.
19. A student focuses on the screen the image of a distant blue-coloured building using a convex lens. The correct characteristics of the image formed are:
(A) Virtual, erect, diminished and in green shade
(B) Real, inverted, diminished and in violet shade
(C) Real, inverted, diminished and in blue shade
(D) Virtual, inverted, diminished and in blue shade
View Answer...
Explanation:
A convex lens forms a real, inverted, and highly diminished image at its focus when the object is very far away (like a distant building). Also, the lens does not change the colour of the object. So a blue building will form a blue image.
20. Media with refractive indices: X=1.51, Y=1.72, Z=1.83, W=2.42. The speed of light is minimum and maximum in:
(A) X–max, W–min
(B) X–min, W–max
(C) Z–min, W–max
(D) X–min, Z–max
View Answer...
We know:
$\mu = \frac{\text{Speed of light in air or vacuum }(c)}{\text{Speed of light in the medium }(v)}$
So,
- larger refractive index → smaller speed
- smaller refractive index → greater speed
Given:
- X = 1.51
- Y = 1.72
- Z = 1.83
- W = 2.42
So:
- maximum speed in X
- minimum speed in W
Hence, (A) X–max, W–min.
21. Match for convex lens (object position → image position):
| 1. At infinity | A. Between F₂ and 2F₂ |
| 2. Beyond 2F₁ | B. At 2F₂ |
| 3. At 2F₁ | C. At infinity |
| 4. At F₁ | D. At F₂ |
(A) 1→D, 2→C, 3→A, 4→B
(B) 1→D, 2→A, 3→B, 4→C
(C) 1→C, 2→D, 3→A, 4→B
(D) 1→D, 2→C, 3→B, 4→A
View Answer...
Explanation: Let’s trace the correct image positions for a convex lens:
Object at infinity $\rightarrow$ Image forms at $F_2$ (Highly diminished) $\rightarrow$ D
Object beyond $2F_1$ $\rightarrow$ Image forms between $F_2$ and $2F_2$ (Diminished) $\rightarrow$ A
Object at $2F_1$ $\rightarrow$ Image forms at $2F_2$ (Same size) $\rightarrow$ B
Object at $F_1$ $\rightarrow$ Image forms at infinity (Highly enlarged) $\rightarrow$ C
22. If the power of a lens is -2D then the type of lens is:
(A) Convex lens
(B) Concave lens
(C) Spherical lens
(D) Both (A) and (B)
View Answer...
Explanation:
Power of a concave lens is always negative, while power of a convex lens is positive.
Given power = -2D, therefore, the lens is a concave lens.
23. If the radius of curvature of a convex mirror is 32 cm, the focal length of the mirror is:
(A) 15 cm
(B) 32 cm
(C) 16 cm
(D) 34 cm
View Answer...
Explanation: The focal length ($f$) of a spherical mirror is:
$f = \frac{R}{2}$
Given $R = 32\text{ cm}$, so, $f = \frac{32}{2} = 16\text{ cm}$.
24. The power of a lens is 1D. Its focal length is:
(A) 1 cm
(B) 0.5 m
(C) 1 m
(D) 0.25 m
View Answer...
Explanation:
- We know, power : $P = \frac{1}{f\text{ (in meters)}}$.
So, $f = \frac{1}{P}$
Given $P = 1\text{ D}$,
So $f = \frac{1}{1} = 1\text{ meter}$.
25. A negative magnification indicates that the image is:
(A) Real image, inverted
(B) Virtual image, erect
(C) Real image, erect
(D) Virtual image, inverted
View Answer...
Explanation:
Magnification ($m$) tells us about the nature of the image. A negative sign ($-$) means the image is real and inverted. A positive sign ($+$) means the image is virtual and erect.
26. Image formed by a concave mirror, when object is placed between F and 2F is:
(A) Real and enlarged
(B) Real and diminished
(C) Virtual and diminished
(D) Virtual and enlarged
View Answer...
Explanation:
For a concave mirror, if the object is placed between F and C (or 2F), the image formed is: real, inverted, and enlarged than the object.
27. The focal length of a lens with power 1D is:
(A) 100 cm
(B) -100 cm
(C) 1 cm
(D) -1 cm
View Answer...
Explanation:
We know that focal length $f = \frac{1}{P}$.
Given, Power ($P$) = $1\text{ D}$
So, $f = \frac{1}{1} = 1\text{ meter}$.
Since 1 m = 100 cm, the focal length is 100 cm.
28. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be:
(A) Plane
(B) Concave
(C) Convex
(D) Either plane or convex
View Answer...
Explanation:
A plane mirror always forms an erect image.
A convex mirror also always forms an erect image.
So, if the image is always erect, the mirror may be plane or convex.
29. An object is placed at a distance of 40 cm in front of a concave mirror of focal length 20 cm. The image formed is:
(A) Real, inverted and same in size
(B) Real, inverted and diminished
(C) Virtual, inverted and enlarged
(D) Virtual, erect and diminished
View Answer...
Explanation:
The focal length ($f$) is $20\text{ cm}$, which means the Centre of Curvature ($C = 2f$) is at $2 \times 20 = 40\text{ cm}$. Since the object is placed exactly at $40\text{ cm}$, it is sitting right on $C$. A concave mirror always forms a real, inverted, and same-sized image when the object is at $C$.
30. If the focal length of a convex lens is 25 cm, then its power will be:
(A) +4D
(B) -4D
(C) +2.5D
(D) -2.5D
View Answer...
Explanation:
Power of lens $P = \frac{1}{f\text{ (in meters)}}$.
Here, $f=25\text{ cm} = 0.25\text{ m}$.
A convex lens has a positive focal length, so $f = +0.25\text{ m}$.
$P = \frac{1}{0.25} = +4\text{ D}$.
31. An object is placed at the centre of curvature of a concave mirror. The distance between its image and the pole is:
(A) Equal to f
(B) Between f and 2f
(C) Equal to 2f
(D) Greater than 2f
View Answer...
Explanation:
The centre of curvature is located at a distance of $2f$ from the pole. When an object is placed at the centre of curvature ($C$), its image is also formed exactly at the centre of curvature ($C$). Therefore, the distance between the image and the pole is equal to $2f$.
32. The radius of curvature of a spherical mirror is 18 cm. What is its focal length?
(A) 9 cm
(B) 36 cm
(C) 9 cm
(D) 0.9 cm
View Answer...
Explanation:
We know, $f = \frac{R}{2}$
Given $R = 18\text{ cm}$
So, $f = \frac{18}{2} = 9\text{ cm}$.
33. Image formed by a convex spherical mirror is:
(A) Virtual
(B) Enlarged
(C) Real
(D) Inverted
View Answer...
Explanation:
A convex mirror always forms an image that is: always virtual, erect, and diminished.
34. The magnification of an image formed by a mirror is m = -1. Which one is true for the mirror:
(A) Only plane mirror
(B) Plane or concave mirror
(C) Convex mirror
(D) Only concave mirror
View Answer...
Explanation:
The negative sign ($-$) of $m $ means the image is real and inverted.
The value $1$ tells us that the image is the exact same size as the object.
Only a concave mirror can form a real, inverted image.
On the other hand, plane mirrors only form virtual images ($m = +1$) and convex mirrors only form diminished virtual images ($m < +1$).
35. Assertion: Convex mirrors are used to look back side of a vehicle.
Reason: Convex mirror formed virtual image.
(A) Assertion and reason both are incorrect
(B) Assertion is correct but Reason is not exact explanation
(C) Assertion is correct and reason is exact explanation
(D) Assertion is incorrect and reason is not exact explanation
View Answer...
Explanation:
Assertion is correct. Convex mirrors are used as rear-view mirrors in vehicles.
Reason is also true because a convex mirror forms a virtual image.
But why is it not the exact explanation?
The main reason we use convex mirrors in vehicles is not just because the image is virtual, but because they form erect, diminished images which gives the driver a much wider field of view to see a large area of traffic behind them.
36. The position of an image formed by a convex lens of focal length 10 cm where the object is placed at a distance of 20 cm:
(A) 20 cm
(B) -20 cm
(C) 20/3 cm
(D) -20/3 cm
View Answer...
Explanation:
Since the focal length ($f$) is $10\text{ cm}$, twice the focal length ($2F$) is $2 \times 10 = 20\text{ cm}$. The object is placed exactly at $20\text{ cm}$ (which is at $2F$). For a convex lens, when an object is placed at $2F_1$, its image is formed on the other side exactly at $2F_2$. Therefore, the image position is $20\text{ cm}$ from the lens.
37. The mirror used as rearview in car is:
(A) Concave
(B) Convex
(C) Plane
(D) None of the above
View Answer...
Explanation:
Convex mirrors are used as rear-view mirrors because they always form an erect (upright) and diminished (smaller) image. This allows drivers to see a much wider area of the traffic behind them compared to plane or concave mirrors.
38. The magnification of plane mirror is:
(A) 1.0
(B) 0.0
(C) 2.0
(D) 3.0
View Answer...
Explanation:
A plane mirror always forms an image that is the exact same size as the object. Since magnification ($m$) is the ratio of image height to object height, $m = 1.0$. The value is positive because the image is always virtual and erect.
39. When the object is placed at the focus of a concave lens, the image formed is:
(A) Real and smaller
(B) Virtual and smaller
(C) Virtual and inverted
(D) Real and erect
View Answer...
Explanation:
A concave lens always forms a virtual, erect, and diminished (smaller) image, regardless of where the object is placed in front of it. It can never form a real or inverted image.
40. When light travels from air to glass:
(A) Angle of incidence > angle of refraction
(B) Angle of incidence < angle of refraction
(C) Angle of incidence = angle of refraction
(D) Not sure
View Answer...
Explanation:
Air is a rarer medium and glass is a denser medium. When light travels from a rarer to a denser medium, it slows down and bends towards the normal line. Because it bends inward, the angle of refraction ($r$) becomes smaller than the angle of incidence ($i$).
41. The velocity of light is maximum in:
(A) Glass
(B) Water
(C) Vacuum
(D) Diamond
View Answer...
Explanation:
Light travels fastest where there are no particles to obstruct its path. The speed of light reaches its maximum value of approximately $3 \times 10^8\text{ m/s}$ in a vacuum. In glass, water, and diamond, the medium slows the light down.
42. The scattering of light by colloidal particles is called:
(A) Tyndall effect
(B) Dispersion
(C) Atmospheric refraction
(D) Internal reflection
View Answer...
Explanation: When a beam of light passes through a colloidal solution, the tiny suspended particles scatter the light rays, making the path of the beam clearly visible. This phenomenon is known as the Tyndall effect (like seeing a beam of sunlight enter a dusty room).
43. For refraction through glass, refractive index μ =
(A) speed of light in water / speed of light in glass
(B) speed of light in glass / speed of light in air
(C) speed of light in water / speed of light in air
(D) speed of light in air / speed of light in glass
View Answer...
Explanation:
Refractive index of a medium is:
$\mu = \frac{\text{Speed of light in air or vacuum }(c)}{\text{Speed of light in the medium }(v)}$
So for glass, $\mu = \frac{\text{Speed of light in air }(c)}{\text{Speed of light in the glass}(v)}$
44. An object at 15 cm is moved towards a convex mirror. Nature of image:
(A) Real and inverted
(B) Real and enlarged
(C) Virtual and diminished
(D) Virtual and enlarged
View Answer...
Explanation: A convex mirror always forms a virtual, erect, and diminished image, no matter where the object is placed. This is true for all positions of the object.
45. When a pencil is partially dipped in water, it appears bent at the water surface because –
(A) Speed of light is maximum in water
(B) Light changes direction on entering a different medium
(C) Pencil floats in water
(D) Intensity of light decreases in water
View Answer...
Explanation:
This is due to the phenomenon of refraction. When light rays travel from water (denser medium) to air (rarer medium), they speed up and bend away from the normal line. This change in direction makes the submerged part of the pencil appear bent to our eyes.
46. Why do dentist use concave mirror to examine a small cavity?
(A) To get diminished and real image
(B) To get enlarged and real image
(C) To get enlarged and virtual image
(D) To get diminished and virtual image
View Answer...
Explanation:
When the object is kept close to a concave mirror, that is, between pole and focus, the mirror forms an image that is: virtual, erect and enlarged. Therefore, dentists use a concave mirror very close to a patient’s tooth.
47. Position for an object in front of concave mirror to get magnified, virtual, erect image –
(A) At F
(B) Within F
(C) Between F and C
(D) At C
View Answer...
Explanation:
A concave mirror forms a virtual, erect and magnified image when the object is placed between the pole and focus.
That means the object should be within F.
48. Refractive indices of media A, B, C, D are 1.47, 1.33, 1.50, 1.36. Light will travel fastest in –
(A) A
(B) B
(C) C
(D) D
View Answer...
Explanation: Light travels fastest in the medium that has the lowest refractive index . Comparing the given values ($1.47, 1.33, 1.50, 1.36$), medium B has the lowest value ($1.33$), so light speeds through it the fastest.
49. Which material cannot be used to make a lens?
(A) Water
(B) Glass
(C) Plastic
(D) Clay
View Answer...
Explanation: To make a lens, a material must be transparent so that light can pass through it and bend (refract). Water, glass, and clear plastics are transparent. Clay is opaque, meaning light cannot pass through it at all.
50. The nature of image formed in a plane mirror is:
(A) Real
(B) Virtual
(C) Virtual, upright and equal
(D) None of the above
View Answer...
Explanation:
The plane mirror always forms an image that is virtual, erect, and equal in size to the actual object.
51. Mirror used to view full length image of a distant tall building –
(A) Concave
(B) Convex
(C) Plane
(D) Concave and plane
View Answer...
Explanation:
A convex mirror gives a wide field of view.
So it helps to see a larger area or a tall object in a smaller image.
That is why it is suitable for viewing the full length image of a distant tall building.
52. A student conducts an activity using an object of height 15 cm in front of a concave mirror. He finds that the image formed is 45 cm in height. What is the magnification of the image?
(A) –3
(B) –1/3
(C) 1/3
(D) 3
Explanation:
We know, Magnification ($m$) is calculated by dividing the height of the image ($h’$) by the height of the object ($h$).
$m = \frac{h’}{h}$
Given $h’ = 45\text{ cm}$
$h = 15\text{ cm}$,
So, $m = \frac{45}{15} = 3$.
For a concave mirror, real image is inverted, so magnification is negative.
So, $m = – 3$.
53. Assertion (A): Virtual images are always erect.
Reason (R): Virtual images are formed by both converging and diverging lens or mirror.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Explanation:
Assertion is true: Virtual images are always upright (erect).
Reason is true: Both converging elements (like a concave mirror or convex lens) and diverging elements (like a convex mirror or concave lens) can form virtual images under different conditions.
But, Reason is not the correct explanation of A. Reason does not provide the underlying physical explanation for why those virtual images must always be erect
54. The relation between radius of curvature (R) and focal length (f) of a spherical mirror is:
(A) R = 1/2 f
(B) R = 2/4 f
(C) R = 16/8 f
(D) R = 16/4 f
Explanation:
The standard formula for the radius of curvature ($R$) and focal length ($f$) is $R = 2f$.
In option (c) : $\frac{16}{8} = 2$. Therefore, $R = \frac{16}{8}f$ is mathematically identical to $R = 2f$.
55. The magnification of a car’s rear view mirror is—
(A) Less than 1
(B) Equal to 1
(C) Greater than 1
(D) None of the above
Explanation: A car’s rear-view mirror is a convex mirror. It is designed to always show a diminished (smaller) image of the traffic behind you. Whenever the image is smaller than the actual object, the magnification value is always less than 1.
56. The angle of refraction will be greater than the angle of incidence when light rays enter from—
(A) Water to glass
(B) Glass to diamond
(C) Air to water
(D) Glass to air
Explanation:
When light travels from denser to rarer medium, it bends away from the normal.
So: angle of refraction > angle of incidence
This happens in glass → air.
57. The bulb of a flashlight or car headlight is placed at the reflector’s—
(A) Focus
(B) Outside the centre of curvature
(C) In between focus and centre of curvature
(D) In between focus and pole
Explanation:
In flashlights and headlights, the bulb is placed exactly at the principal focus of a concave reflector. According to the laws of reflection, any light rays originating from the focus will strike the concave surface and bounce back as a powerful, parallel beam of light that travels a long distance.
58. Assertion (A): A –2.5D powered lens is a converging lens.
Reason (R): The power of a lens is defined as the reciprocal of its focal length.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Explanation:
Assertion is false: A lens with a negative power ($-2.5\text{ D}$) has a negative focal length, which means it is a concave (diverging) lens, not a converging one.
Reason is true: The definition of power is indeed the reciprocal of its focal length in meters ($P = \frac{1}{f}$)
59. Match the medium with its refractive index:
| Column A (Medium) | Column B (Refractive Index) |
| (i) Air | (p) 1.36 |
| (ii) Water | (q) 2.42 |
| (iii) Diamond | (r) 1.33 |
| (iv) Alcohol | (s) 1.0003 |
(A) (i)→s, (ii)→r, (iii)→q, (iv)→p
(B) (i)→s, (ii)→r, (iii)→p, (iv)→q
(C) (i)→s, (ii)→p, (iii)→q, (iv)→r
(D) (i)→p, (ii)→q, (iii)→s, (iv)→r
Explanation:
Air has the lowest index $\rightarrow$ $1.0003$ (s)
Water $\rightarrow$ $1.33$ (r)
Alcohol $\rightarrow$ $1.36$ (p)
Diamond has the highest index $\rightarrow$ $2.42$ (q)
60. A concave mirror of focal length 20 cm forms an image having twice the size of object. For the virtual image the position of the object will be at:
(A) 25 cm
(B) 40 cm
(C) 10 cm
(D) Infinity
Explanation:
For a concave mirror, a virtual image is always formed when the object is placed inside the focus (between $P$ and $F$). Since the focal length is $20\text{ cm}$, the object must be at a distance less than $20\text{ cm}$. Out of the options given, only 10 cm satisfies this condition.
61. Whatever be the position of the object, the image formed by a mirror is virtual, erect and smaller than the object. Then the mirror must be:
(A) Plane
(B) Concave
(C) Convex
(D) Either concave or convex
Explanation:
No matter where you place an object in front of it, a convex mirror always forms a virtual, erect, and smaller (diminished) image behind the mirror.
62. In an activity Partha holds a convex mirror in one hand and on the other hand holds a pen in upright position. He observes the image of the pencil in the mirror. Now he moves the pen away from the mirror slowly and repeats this activity carefully. Select the correct observation when the pen is moved away from the mirror slowly:
(i) The image becomes smaller
(ii) The image becomes larger
(iii) The image moves from the pole towards the focus
(iv) The image moves farther away from the focus
(A) Both (i) and (iii)
(B) Both (ii) and (iii)
(C) Both (ii) and (iii)
(D) Both (ii) and (iv)
Explanation:
For a convex mirror:
- the image is always virtual, erect and diminished
- as the object moves away, the image becomes smaller
- the image shifts towards the focus
So the correct observations are:
(i) The image becomes smaller
(iii) The image moves from the pole towards the focus
Hence, the answer is (A).
63. Arpita is standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body of the same size and that of the legs smaller. The following is the order of combinations for the magic mirror from the top:
(A) Plane, convex and concave
(B) Convex, concave and plane
(C) Concave, plane and convex
(D) Convex, plane and concave
Explanation:
Head bigger → needs a concave mirror because it can form a magnified image
Middle same size → plane mirror
Legs smaller → convex mirror because it forms diminished image
So the order is:
Concave → Plane → Convex
64. The speed of light in air is 3 × 10⁸ ms⁻¹ and in medium X is 1.5 × 10⁸ ms⁻¹. The refractive index of X will be:
(A) 2
(B) 4.5
(C) 0.5
(D) 1.5
Explanation:
We know that, the refractive index of a medium, $n = \frac{\text{Speed of light in air}}{\text{Speed of light in the medium}}$.
$n = \frac{3 \times 10^8\text{ ms}^{-1}}{1.5 \times 10^8\text{ ms}^{-1}}$
$n = \frac{3}{1.5} = 2$.
65. A safety mirror placed at a roadside has a radius of curvature of 50 cm. If a person standing at a distance of 100 cm from the safety mirror, what will be the position of the image formed in the safety mirror?
(A) 25 cm
(B) 20 cm
(C) 2.5 cm
(D) 50 cm
Explanation: A roadside safety mirror is always a convex mirror.
Given, Radius of curvature ($R$) = $+50\text{ cm}$
So focal length ($f$) = $\frac{50}{2} = +25\text{ cm}$.
Object distance ($u$) = $-100\text{ cm}$.
We know that: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$ \implies \frac{1}{v} + \frac{1}{-100} = \frac{1}{25}$
$\frac{1}{v} = \frac{1}{25} + \frac{1}{100}$
$\frac{1}{v}= \frac{4 + 1}{100}$
$\frac{1}{v}= \frac{5}{100} = \frac{1}{20}$
Therefore, the image forms at $20\text{ cm}$ behind the mirror.
66. The correct decreasing order of optical density of transparent medium is:
(A) Air > Ice > Alcohol > Glass
(B) Glass > Ice > Alcohol > Air
(C) Glass > Alcohol > Ice > Air
(D) Glass > Ice > Air > Alcohol
Explanation:
Greater refractive index means greater optical density.
Approximate refractive indices:
- Glass ≈ 1.5
- Alcohol ≈ 1.36
- Ice ≈ 1.31
- Air ≈ 1.0003
So decreasing order of optical density is: (C) Glass > Alcohol > Ice > Airc
67. Virtual, erect and diminished image is formed:
(i) Convex mirror
(ii) Concave mirror
(iii) Convex lens
(iv) Concave lens
(A) i, ii
(B) ii, iii
(C) i, iv
(D) ii, iv
Explanation:
(i) Convex mirror → virtual, erect, diminished ✔
(ii) Concave mirror → can form virtual erect image, but it is magnified, not diminished ✘
(iii) Convex lens → when virtual image is formed, it is magnified, not diminished ✘
(iv) Concave lens → virtual, erect, diminished ✔
So the correct answer is (C) (i) and (iv).
68. Assertion (A): The S.I. Unit of power of a lens is diopter.
Reason (R): The power of a concave lens is positive and that of a convex lens is negative.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Explanation:
Assertion is true: The S.I. unit of lens power is the dioptre (D).
Reason is false: By sign convention, a concave lens has a negative focal length (so its power is negative), while a convex lens has a positive focal length (so its power is positive).
69. Assertion: Magnification of a mirror has no unit.
Reason: The ratio of height of the image to the height of the object is the linear magnification produced by mirror.
(A) Both A and B are true and R is correct explanation of the assertion
(B) Both A and B are true R is not correct explanation of the assertion
(C) A is true but R is false
(D) A is false but R is true
Explanation:
Magnification is calculated by dividing the height of the image by the height of the object ($m = \frac{h’}{h}$).
It is a ratio of two lengths, so units cancel out.
Therefore:
- Assertion is true
- Reason is also true
- Reason correctly explains the assertion
70. The power of a plane mirror is –
(A) 0
(B) 1
(C) –1
(D) ∞
Explanation:
A plane mirror has an infinite focal length ($f = \infty$).
So, power is : $P = \frac{1}{f} = \frac{1}{\infty} = 0$.
71. The magnification produced by a convex lens –
(A) |m| = 1
(B) |m| > 1
(C) |m| < 1
(D) Either |m| > 1 or |m| < 1
Explanation:
A convex lens can form different types of images depending on object position.
So magnification may be:
- greater than 1 → enlarged image
- less than 1 → diminished image
- even equal to 1 when object is at 2F
Among the given options, the best answer is (D).
72. For a spherical mirror, to get a highly diminished image, the object should be placed –
(A) At focus
(B) At infinity
(C) Beyond centre of curvature
(D) Between principal focus and pole
Explanation:
When an object is placed at infinity in front of a concave mirror, the image formed is highly diminished and forms at the focus.
73. The image formed by concave lenses is –
(A) Real, diminished
(B) Virtual, diminished
(C) Virtual, same sized
(D) Virtual, enlarged
Explanation:
A concave lens always forms an virtual, erect, and smaller (diminished) than the object.
74. A concave mirror has focal length 15cm. It produce virtual and erect image. Then position of the object be
(A) At the distance 15 cm from mirror
(B) At the distance 30 cm from mirror
(C) At the distance 10 cm from mirror
(D) At the distance 40 cm from mirror
Explanation: A concave mirror only produces a virtual and erect image when the object is placed very close to it—specifically inside its focal length (between $P$ and $F$). Since the focal length is $15\text{ cm}$, the object must be placed at a distance less than $15\text{ cm}$. Only 10 cm fits this rule.