Geometrical Optics | Physics JEE previous year questions with Solutions

Q 1 :
A convex mirror of radius of curvature 30 cm forms an image that is half the size of the object. The object distance is [2024]

-15 cm
45 cm
-45 cm
15 cm

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(1)

Given R = 30 cm

$\Rightarrow f = \frac{R}{2} = + 15 c m$

Magnification (m) = $\pm 1 / 2$

For convex mirror, virtual image is formed for real object, therefore, $m$ is +ve

The mirror formula,

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow v = \frac{u f}{u - f}$

Magnification, $m = - \frac{v}{u} = - \frac{u f}{u (u - f)} = \frac{f}{f - u}$

$\frac{1}{2} = \frac{f}{f - u} = \frac{15}{15 - u} \Rightarrow u = - 15 c m$

Q 2 :
If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be [2024]

10/3 cm
-20 cm
-10 cm
15 cm

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(3)

Magnification, $m = 2 = \frac{- v}{u}$

$2 = \frac{- (15 - u)}{- u}$

$2 u = 15 - u$

$3 u = 15 \Rightarrow u = 5 cm$

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{10} + \frac{1}{- 5} = \frac{1 - 2}{10} = \frac{- 1}{10}$

$f = - 10 cm$

Q 3 :
A concave mirror of focal length f in air is dipped in a liquid of refractive index $μ$ . Its focal length in the liquid will be [2025]

$\frac{f}{μ}$
$\frac{f}{(μ - 1)}$
$μ f$
f

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(4)

Focal length of mirror will not change because focal length of mirror dosen't depend on medium.

Q 4 :
A concave mirror produces an image of an object such that the distance between the object and image is 20 cm. If the magnification of the image is –3, then the magnitude of the radius of curvature of the mirror is: [2025]

3.75 cm
30 cm
7.5 cm
15 cm

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(4)

$m = - 3 = - \frac{v}{u}$

u = –x

v = –3x

$\Rightarrow$ 2x = 20 cm

x = 10 cm

$f = \frac{u v}{u + v} = \frac{(- 10) (- 30)}{(- 10) + (- 30)} = - 7.5 cm$

$f = - \frac{R}{2}$

R = –2f = 15 cm

Q 5 :
Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of 12 cm, at same distance of 18 cm from the respective mirrors. The ratio of sizes of the images formed by convex mirror and by concave mirror is: [2025]

1/2
2
3
1/3

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4

(1)

In case of concave mirror,

Using $m = \frac{f}{u - f}$

$m_{1} = \frac{6}{18 - 6} = \frac{1}{2}$

In case of convex mirror,

$m_{2} = \frac{6}{18 + 6} = \frac{1}{4} ∴ \frac{m_{2}}{m_{1}} = \frac{1}{2}$

Q 6 :
When an object is placed 40 cm away from a spherical mirror an image of magnification $\frac{1}{2}$ is produced. To obtain an image with magnification of $\frac{1}{3}$ , the object is to be moved. [2025]

40 cm away from the mirror.
80 cm away from the mirror.
20 cm towards the mirror.
20 cm away from the mirror.

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(1)

$m = \frac{1}{2} = \frac{f}{f - u} n$

$\frac{1}{2} = \frac{f}{f - (- 40)}$

$f + 40 = 2 f \Rightarrow f = 40 cm$

now $m = \frac{1}{3} = \frac{40}{40 - u}$

$40 - u = 120 \Rightarrow u = - 80 cm$

Move 40 cm away from mirror.

Q 7 :
A finite size object is placed normal to the principal axis at a distance of 30 cm from a convex mirror of focal length 30 cm. A plane mirror is now placed in such a way that the image produced by both the mirrors coincide with each other. The distance between the two mirrors is: [2025]

45 cm
7.5 cm
22.5 cm
15 cm

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(2)

For Convex mirror : $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} - \frac{1}{30} = \frac{1}{30}$

$\frac{1}{v} = \frac{2}{30} = \frac{1}{15} \Rightarrow v = 15 cm$

Image formed by convex mirror is at 45 cm from object so plane mirror should be placed midway at 22.5 cm from object so that both of their images may coincide.

Therefore, distance between both mirrors,

d = 30 – 22.5 = 7.5 cm

Q 8 :
A mirror is used to produce an image with magnification of $\frac{1}{4}$ . If the distance between object and its image is 40 cm, then the focal length of the mirror is [2025]

10 cm
12.7 cm
10.7 cm
15 cm

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(3)

$m = - \frac{v}{u} = - (\frac{v}{- u}) = \frac{v}{u}$

$\frac{1}{4} = \frac{v}{u} \Rightarrow u = 4 v$

$v + u = 40 \Rightarrow 5 v = 40$

v = 8 cm and u = 32 cm

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{8} - \frac{1}{32} = \frac{1}{f}$

$\Rightarrow f = \frac{32}{3} = 10.7 cm$

Q 9 :
The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature R = 2 m. Another car approaches him from behind with a uniform speed of 90 km/hr. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the side view mirror is 'a'. The value of 100a is ________ $m / s^{2}$ . [2025]

0%

(8)

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} + (\frac{1}{- 24}) = 1 \Rightarrow v = \frac{24}{25} cm$

$\Rightarrow m = - \frac{v}{u} = - \frac{24}{25 (- 24)} = \frac{1}{25}$

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ ... (i)

Diff. (i) w.r.t. time

$\frac{- 1}{v^{2}} (\frac{d v}{d t}) - \frac{1}{u^{2}} (\frac{d u}{d t}) = 0$ ... (ii)

$V_{1} = - m^{2} V_{0} = - {(\frac{1}{25})}^{2} \cdot 25 = - \frac{1}{25} {ms}^{- 1}$

Diff. (ii) w.r.t. time

$- (- \frac{2}{v^{3}} \cdot {(V_{I})}^{2} + \frac{1}{v^{2}} \cdot a_{I}) - (\frac{2}{u^{3}} \cdot {(V_{O})}^{2} + \frac{1}{u^{2}} \cdot a_{O}) = 0$

$\frac{a_{I}}{v^{2}} = \frac{2}{v^{3}} \cdot V_{I}^{2} - \frac{2}{u^{3}} \cdot V_{O}^{2}$

$\Rightarrow a_{I} = \frac{2}{v} \cdot V_{I}^{2} - \frac{2 v^{2}}{u^{3}} \cdot V_{O}^{2}$

$\Rightarrow a_{I} = \frac{2}{(\frac{24}{25})} \cdot {(\frac{1}{25})}^{2} - \frac{2 {(\frac{24}{25})}^{2}}{{(24)}^{2}} \cdot {(25)}^{2} = \frac{- 2}{25} {ms}^{- 2}$

$| 100 a_{I} |$ $= \frac{2}{25} \times 100 = 8$

Q 10 :
Distance between object and its image (magnified by $- \frac{1}{3}$ ) is 30 cm. The focal length of the mirror used is $(\frac{x}{4}) cm$ , where magnitude of value of x is __________. [2025]

0%

(45)

$m = \frac{f}{f - u} = - \frac{1}{3} \Rightarrow 4 f = u$

$\frac{1}{v} + \frac{1}{4 f} = \frac{1}{f} \Rightarrow v = \frac{4 f}{3}$

$(u - v) = \frac{8 f}{3} = 30$

$f = \frac{90}{8} = \frac{45}{4} \Rightarrow x = 45$

Topic Question Set

Q 1 :
A convex mirror of radius of curvature 30 cm forms an image that is half the size of the object. The object distance is [2024]

Q 2 :
If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be [2024]

Q 3 :
A concave mirror of focal length f in air is dipped in a liquid of refractive index $μ$ . Its focal length in the liquid will be [2025]

Q 4 :
A concave mirror produces an image of an object such that the distance between the object and image is 20 cm. If the magnification of the image is –3, then the magnitude of the radius of curvature of the mirror is: [2025]

Q 5 :
Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of 12 cm, at same distance of 18 cm from the respective mirrors. The ratio of sizes of the images formed by convex mirror and by concave mirror is: [2025]

Q 6 :
When an object is placed 40 cm away from a spherical mirror an image of magnification $\frac{1}{2}$ is produced. To obtain an image with magnification of $\frac{1}{3}$ , the object is to be moved. [2025]

Q 7 :
A finite size object is placed normal to the principal axis at a distance of 30 cm from a convex mirror of focal length 30 cm. A plane mirror is now placed in such a way that the image produced by both the mirrors coincide with each other. The distance between the two mirrors is: [2025]

Q 8 :
A mirror is used to produce an image with magnification of $\frac{1}{4}$ . If the distance between object and its image is 40 cm, then the focal length of the mirror is [2025]

0%

Q 10 :
Distance between object and its image (magnified by $- \frac{1}{3}$ ) is 30 cm. The focal length of the mirror used is $(\frac{x}{4}) cm$ , where magnitude of value of x is __________. [2025]

0%

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Topic Question Set

Q 1 : A convex mirror of radius of curvature 30 cm forms an image that is half the size of the object. The object distance is [2024]

Q 2 : If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be [2024]

Q 3 : A concave mirror of focal length f in air is dipped in a liquid of refractive index μ. Its focal length in the liquid will be [2025]

Q 4 : A concave mirror produces an image of an object such that the distance between the object and image is 20 cm. If the magnification of the image is –3, then the magnitude of the radius of curvature of the mirror is: [2025]

Q 5 : Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of 12 cm, at same distance of 18 cm from the respective mirrors. The ratio of sizes of the images formed by convex mirror and by concave mirror is: [2025]

Q 6 : When an object is placed 40 cm away from a spherical mirror an image of magnification 12 is produced. To obtain an image with magnification of 13, the object is to be moved. [2025]

Q 7 : A finite size object is placed normal to the principal axis at a distance of 30 cm from a convex mirror of focal length 30 cm. A plane mirror is now placed in such a way that the image produced by both the mirrors coincide with each other. The distance between the two mirrors is: [2025]

Q 8 : A mirror is used to produce an image with magnification of 14. If the distance between object and its image is 40 cm, then the focal length of the mirror is [2025]

0%

Q 10 : Distance between object and its image (magnified by –13) is 30 cm. The focal length of the mirror used is (x4) cm, where magnitude of value of x is __________. [2025]

0%

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Q 1 :
A convex mirror of radius of curvature 30 cm forms an image that is half the size of the object. The object distance is [2024]

Q 2 :
If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be [2024]

Q 3 :
A concave mirror of focal length f in air is dipped in a liquid of refractive index $μ$ . Its focal length in the liquid will be [2025]

Q 4 :
A concave mirror produces an image of an object such that the distance between the object and image is 20 cm. If the magnification of the image is –3, then the magnitude of the radius of curvature of the mirror is: [2025]

Q 5 :
Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of 12 cm, at same distance of 18 cm from the respective mirrors. The ratio of sizes of the images formed by convex mirror and by concave mirror is: [2025]

Q 6 :
When an object is placed 40 cm away from a spherical mirror an image of magnification $\frac{1}{2}$ is produced. To obtain an image with magnification of $\frac{1}{3}$ , the object is to be moved. [2025]

Q 7 :
A finite size object is placed normal to the principal axis at a distance of 30 cm from a convex mirror of focal length 30 cm. A plane mirror is now placed in such a way that the image produced by both the mirrors coincide with each other. The distance between the two mirrors is: [2025]

Q 8 :
A mirror is used to produce an image with magnification of $\frac{1}{4}$ . If the distance between object and its image is 40 cm, then the focal length of the mirror is [2025]

Q 10 :
Distance between object and its image (magnified by $- \frac{1}{3}$ ) is 30 cm. The focal length of the mirror used is $(\frac{x}{4}) cm$ , where magnitude of value of x is __________. [2025]