Geometrical Optics — Page 11 | JEE Physics

Q101

When an object is placed 40 cm away from a spherical mirror an image of magnification

\dfrac{1}{2}

is produced. To obtain an image with magnification of

\dfrac{1}{3}

, the object is to be moved :

A 20 cm away from the mirror.

B 20 cm towards the mirror.

C 80 cm away from the mirror.

D 40 cm away from the mirror.

Correct Answer

Option D

Solution

To solve this problem, start by using the magnification formula for spherical mirrors: $m = \dfrac{f}{f - u}$ Given that the initial magnification $m = \dfrac{1}{2}$ , and the object distance $u = -40$ cm (the negative sign indicates distance measured against the incoming light), we can write: $\dfrac{1}{2} = \dfrac{f}{f - (-40)}$ This simplifies to: $\dfrac{1}{2} = \dfrac{f}{f + 40}$ Cross-multiplying gives: $f + 40 = 2f$ This leads to: $f = 40 \text{ cm}$ To find the new object distance for a magnification of $m = \dfrac{1}{3}$ , we use the same formula: $m = \dfrac{f}{f - u}$ Substitute the known values: $\dfrac{1}{3} = \dfrac{40}{40 - u}$ Cross-multiplying gives: $40 - u = 120$ Solving for $u$ results in: $u = -80 \text{ cm}$ Thus, the object needs to be placed 80 cm from the mirror, opposite to the initial placement of 40 cm.

To achieve this repositioning, the object needs to be moved 40 cm further from its initial 40 cm distance, totaling a move of 40 cm away from the mirror.

Q102

The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea (7.8 mm). This surface separateds two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus -

A 2 cm

B 3.1 cm

C 4.0 cm

D 1 cm

Correct Answer

Option B

Solution

{{1.34} \over V} - {1 \over \infty } = {{1.34 - 1} \over {7.8}}

$\therefore$ V = 30.7 mm

Q103

Let the refractive index of a denser medium with respect to a rarer medium be n12 and its critical angle be θC . At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is 90o. Angle A is given by :

A

{1 \over {{{\cos }^{ - 1}}\left( {\sin {\theta _C}} \right)}}

B

{1 \over {{{\tan }^{ - 1}}\left( {\sin {\theta _C}} \right)}}

C

{\cos ^{ - 1}}\,\left( {\sin {\theta _C}} \right)

D

{\tan ^{ - 1}}\,\left( {\sin {\theta _C}} \right)

Correct Answer

Option D

Solution

Refractive index, n12 =

{{{n_D}} \over {{n_R}}}

=

{1 \over {\sin {\theta _c}}}

$\Rightarrow$

\,\,\,

sin $\theta$ c =

{{{n_R}} \over {{n_D}}}

. . . . . (1) From Snell's law, nD sinA = nR sin r

{{{n_R}} \over {{n_D}}}

=

{{\sin A} \over {\sin \left( {{{90}^o} - A} \right)}}

[as r = 90o $-$ A] $\Rightarrow$

\,\,\,

{{{n_R}} \over {{n_D}}} = \tan A

\therefore\,\,\,

From (1) we get, tan A = sin $\theta$ c $\Rightarrow$

\,\,\,

A = tan $-$ 1 (sin $\theta$ c)

Q104

A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is :

A 1.1 cm away from the lens

B 0

C 0.55 cm towards the lens

D 0.55 cm away from the lens

Correct Answer

Option D

Solution

Image is formed on the screen. So, v = 10 cm and $\mu$ = $-$ 10 cm Using formula,

{1 \over v} - {1 \over u} = {1 \over f}

$\Rightarrow$

{1 \over {10}} - {1 \over {\left( { - 10} \right)}}

=

{1 \over f}

$\Rightarrow$ f = 5 cm Now a glass block is placed like this, Because of this glass block source will move t

\left( {1 - {1 \over \mu }} \right)

in the direction of incident ray. $\therefore$ S' = t

\left( {1 - {1 \over \mu }} \right)

= 1.5

\left( {1 - {2 \over 3}} \right)

= 0.5 $\therefore$ now distance of source from the lens = 10 $-$ 0.5 = 9.5 cm $\therefore$ $\mu$ = $-$ 9.5 cm $\therefore$

{1 \over v} - {1 \over {\left( { - 9.5} \right)}}

=

{1 \over 5}

$\Rightarrow$

{1 \over v}

=

{1 \over 5} - {2 \over {19}}

$\Rightarrow$

{1 \over v}

=

{9 \over {95}}

$\Rightarrow$ v = 10.55 cm So, to get sharp image screen should shift away (10.55 $-$ 10) = 0.55 cm from the lens.

Q105

To find the focal length of a convex mirror, a student records the following data : .tg .tg Object Pin Convex Lens Convex Mirror Image Pin 22.2 cm 32.2 cm 45.8 cm 71.2 cm The focal length of the convex lens is f1 and that of mirror is f2. Then taking index correction to be negligibly small, f1 and f2 are close to :

A f1 = 12.7 cm f2 = 7.8 cm

B f1 = 7.8 cm f2 = 12.7 cm

C f1 = 7.8 cm f2 = 25.4 cm

D f1 = 15.6 cm f2 = 25.4 cm

Correct Answer

Option B

Solution

For lens : u1 = $-$ (32.2 $-$ 22.2) cm = $-$ 10 cm v1 = (71.2 $-$ 32.2) cm = 39 cm $\therefore$

{1 \over {{f_1}}}

=

{1 \over {{v_1}}} - {1 \over {{u_1}}}

=

{1 \over {39}}

+

{1 \over {10}}

=

{{49} \over {390}}

$\therefore$ f1 = 7.8 cm For mirror : R = (71.2 $-$ 45.8) cm = 25.4 cm $\therefore$ f2 =

{R \over 2}

=

{{25.4} \over 2}

= 12.7 cm

Q106

Wavelength of light used in an optical instrument are

{\lambda _1} = 4000\mathop A\limits^ \circ

and

{\lambda _2} = 5000\mathop A\limits^ \circ ,

then ratio of their respective resolving powers (corresponding to

{\lambda _1}

and

{\lambda _2}

) is :

A

16:25

B

9:1

C

4:5

D

5:4

Correct Answer

Option D

Solution

The resolving power (RP) of an optical instrument is inversely proportional to the wavelength (λ) of light used.

So, if we denote the resolving powers corresponding to λ₁ and λ₂ as RP₁ and RP₂ respectively, we can express this relationship as : RP $\propto$

{1 \over \lambda }

Therefore, the ratio of the resolving powers corresponding to λ₁ and λ₂ is given by the inverse of the ratio of the wavelengths.

In LaTeX notation, this relationship can be written as : $\dfrac{RP_1}{RP_2} = \dfrac{\lambda_2}{\lambda_1}$ Substituting the given wavelengths into this equation, we get : $\dfrac{RP_1}{RP_2} = \dfrac{5000 \, \text{Å}}{4000 \, \text{Å}} = \dfrac{5}{4}$ So, the correct answer is : Option D : 5 : 4.

Q107

A light wave travelling linearly in a medium of dielectric constant 4, incidents on the horizontal interface separating medium with air. The angle of incidence for which the total intensity of incident wave will be reflected back into the same medium will be : (Given : relative permeability of medium

\mu

r = 1)

A 10

^\circ

B 20

^\circ

C 30

^\circ

D 60

^\circ

Correct Answer

Option D

Solution

n = \sqrt {K\mu } = 2

(n $\Rightarrow$ refractive index) So for TIR

\theta > {\sin ^{ - 1}}\left( {{1 \over n}} \right)

\theta > 30

Only option is 60

^\circ

Q108

The difference of speed of light in the two media A and B (vA

-

vB) is 2.6

\times

107 m/s. If the refractive index of medium B is 1.47, then the ratio of refractive index of medium B to medium A is : (Given : speed of light in vacuum c = 3

\times

108 ms

-

1)

A 1.303

B 1.318

C 1.13

D 0.12

Correct Answer

Option C

Solution

Speed of light in a medium

= {c \over n}

$\Rightarrow$ According to given information,

{c \over {{n_A}}} - {c \over {{n_B}}} = 2.6 \times {10^7}

\Rightarrow {{{n_B}} \over {{n_A}}} - 1 = {{2.6 \times {{10}^7}} \over {3 \times {{10}^8}}} \times {n_B}

\Rightarrow {{{n_B}} \over {{n_A}}} \simeq 1.13

Q109

Which of the following statement is correct?

A In primary rainbow, observer sees red colour on the top and violet on the bottom

B In primary rainbow, observer sees violet colour on the top and red on the bottom

C In primary rainbow, light wave suffers total internal reflection twice before coming out of water drops.

D Primary rainbow is less bright than secondary rainbow.

Correct Answer

Option A

Solution

In primary rainbow, observer sees red colour on the top and violet on the bottom.

Q110

A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. An object is placed at 20 cm to the left of this lens system. The distance of the image from the lens in cm is ________.

A

\dfrac{60}{7}

B 15

C 45

D 30

Correct Answer

Option B

Solution

Equivalent focal length

\begin{aligned} & \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\ & =\frac{1}{30}+\frac{1}{-20}=\frac{2-3}{60}=-\frac{1}{60} \\ & \mathrm{f}=-60 \mathrm{~cm} \end{aligned}

Lens formula

\begin{aligned} & \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ & \frac{1}{\mathrm{v}}-\frac{1}{-20}=\frac{1}{-60} \\ & \mathrm{v}=-15 \mathrm{~cm} \end{aligned}