Geometrical Optics — Page 12 | JEE Physics

Q111

A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be

A

\dfrac{700}{11}

cm

B

\dfrac{600}{11}

cm

C

\dfrac{800}{11}

cm

D

\dfrac{500}{11}

cm

Correct Answer

Option B

Solution

\begin{aligned} \frac{1}{\mathrm{f}} & =\left(\frac{1.3-1}{1}\right)\left(\frac{1}{\infty}-\frac{1}{-30}\right) \\ & =\left(\frac{1.5-1}{1}\right)\left(\frac{1}{-30}-\frac{1}{-30}\right) \\ & =\frac{0.3}{30}+\frac{0.5}{60}=\frac{1}{100}+\frac{1}{120} \\ & =\frac{6+5}{600}=\frac{11}{600} \\ \mathrm{f} & =\frac{600}{11} \mathrm{~cm} \end{aligned}

Q112

A plano convex lens of refractive index

1.5

and radius of curvature

30

cm

. Is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of the object

A

60

cm

B

30

cm

C

20

cm

D

80

cm

Correct Answer

Option C

Solution

KEY CONCEPT : The focal length

\left( F \right)

of the final mirror is

{1 \over F} = {2 \over {f\ell }} + {1 \over {{f_m}}}

Here

{1 \over {{f_\ell }}} = \left( {\mu - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

= \left( {1.5 - 1} \right)\left[ {{1 \over \alpha } - {1 \over { - 30}}} \right] = {1 \over {60}}

$\therefore$

{1 \over F} = 2 \times {1 \over {60}} + {1 \over {30/2}} = {1 \over {10}}

$\therefore$

F=10cm

The combination acts as a converging mirror. For the object to be of the same size of mirror,

u = 2F = 20cm

Q113

A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is

{4 \over 3}

and the fish is

12

cm

below the surface, the radius of this circle in

cm

is

A

{{36} \over {\sqrt 7 }}

B

{36\sqrt 7 }

C

{4\sqrt 5 }

D

{36\sqrt 5 }

Correct Answer

Option A

Solution

\sin {\theta _c} = {1 \over \mu } = {3 \over 4}

or

\tan {\theta _c} = {3 \over {\sqrt {16 - 9} }} = {3 \over {\sqrt 7 }} = {R \over {12}}

\Rightarrow R = {{36} \over {\sqrt 7 }}\,cm

Q114

A double convex lens has power P and same radii of curvature R of both the surfaces. The radius of curvature of a surface of a plano-convex lens made of the same material with power 1.5 P is :

A

{R \over 3}

B

{{3R} \over 2}

C

{R \over 2}

D 2R

Correct Answer

Option A

Solution

Assume refractive index = $\mu$

l

P =

\left( {{\mu _l} - 1} \right)\left( {{2 \over R}} \right)

.....(1)

{3 \over 2}P = \left( {{\mu _l} - 1} \right)\left( {{1 \over {{R_1}}}} \right)

......(2) from (1)/(2)

{P \over {{3 \over 2}P}} = {{\left( {{2 \over R}} \right)} \over {\left( {{1 \over {{R_1}}}} \right)}}

$\Rightarrow$ R1 =

{R \over 3}

Q115

The refractive index of a glass is

1.520

for red light and

1.525

for blue light. Let

{D_1}

and

{D_2}

be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then,

A

{D_1} < {D_2}

B

{D_1} = {D_2}

C

{D_1}

can be less than or greater than

{D_2}

depending upon the angle of prism

D

{D_1} > {D_2}

Correct Answer

Option A

Solution

For a thin prism,

D = \left( {\mu - 1} \right)A

Since

{\lambda _b} < {\lambda _r} \Rightarrow {\mu _r} < {\mu _b} \Rightarrow {D_1} < {D_2}

Q116

Two lenses of power

-15

D

and

+5

D

are in contact with each other. The focal length of the combination is

A

+ 10\,cm

B

- 20\,cm

C

- 10\,cm

D

+ 20\,cm

Correct Answer

Option C

Solution

Power of combination is given by

P = {P_1} + {P_2} = \left( { - 15 + 5} \right)D

\,\,\,\,\,\,\,\,\, = - 10D.

Now,

P = {1 \over f} \Rightarrow f = {1 \over P} = {1 \over { - 10}}

metre $\therefore$

f = - \left( {{1 \over {10}} \times 100} \right)cm = - 10\,cm.

Q117

If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece, should be close to :

A 22 mm

B 12 mm

C 33 mm

D 2 mm

Correct Answer

Option A

Solution

Case 1 : Near – point adjustment M.P =

{L \over {{f_0}}}\left( {1 + {D \over {{f_e}}}} \right)

$\Rightarrow$ 375 =

{{150} \over 5}\left( {1 + {{250} \over {{f_e}}}} \right)

$\Rightarrow$ fe = 21.7 mm $\approx$ 22 mm Case-2 : If final image is at inifinity M.P =

{L \over {{f_0}}}\left( {{D \over {{f_e}}}} \right)

$\Rightarrow$ 375 =

{{150} \over 5}\left( {{{250} \over {{f_e}}}} \right)

$\Rightarrow$ fe = 20 mm

Q118

For the thin convex lens, the radii of curvature are at

15 \mathrm{~cm}

and

30 \mathrm{~cm}

respectively. The focal length the lens is

20 \mathrm{~cm}

. The refractive index of the material is :

A 1.2

B 1.5

C 1.4

D 1.8

Correct Answer

Option B

Solution

To find the refractive index of the material of a thin convex lens, we can make use of the Lensmaker's Formula.

The Lensmaker's formula is given by:

\frac{1}{f} = \left( \frac{\mu - 1}{\mu} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)

where $f$ is the focal length of the lens, $\mu$ is the refractive index of the material of the lens, $R_1$ is the radius of curvature of the first surface (convex side, positive), $R_2$ is the radius of curvature of the second surface (concave side, negative for convex lens).

Given in the question, the radii of curvature are $15 \, \mathrm{cm}$ and $30 \, \mathrm{cm}$ respectively, and the focal length $f = 20\, \mathrm{cm}$ .

It's important to pay attention to the signs of the radii of curvature according to the lens maker's convention.

For convex lenses, $R_1$ is positive and $R_2$ is negative; however, since the problem doesn't specify which curvature corresponds to which side in relation to the direction of light travel, we'll assume the light travels from left to right: thus, $R_1 = +15 \, \mathrm{cm}$ and $R_2 = -30 \, \mathrm{cm}$ .

Substituting the given values into the Lensmaker's Formula, we get:

\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} - \frac{1}{-30} \right)

\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} + \frac{1}{30} \right)

\frac{1}{20} = (\mu - 1) \left( \frac{2 + 1}{30} \right)

\frac{1}{20} = (\mu - 1) \left( \frac{3}{30} \right)

\frac{1}{20} = (\mu - 1) \left( \frac{1}{10} \right)

\frac{1}{20} = \frac{\mu - 1}{10}

Now, solve for $\mu$ :

\mu - 1 = \frac{10}{20}

\mu - 1 = 0.5

\mu = 1.5

Hence, the refractive index of the material of the lens is 1.5, which corresponds to Option B.

Q119

A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is

2

cm. The focal lengths of the component lenses are :

A 10 cm, 12 cm

B 12 cm, 14 cm

C 16 cm, 18 cm

D 18 cm, 20cm

Correct Answer

Option D

Solution

For a convergent doublet of separated lens, we have

{1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {d \over {{f_1}{f_2}}}

...... (1) where d is separation between two lens, f1 and f2 are focal lengths of component lenses, f is resultant focal length.

Therefore, Eq. (1) becomes

{1 \over {10}} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {2 \over {{f_1}{f_2}}} \Rightarrow {1 \over {10}} = \left( {{{{f_2} + {f_1} - 2} \over {{f_1}{f_2}}}} \right)

\Rightarrow {f_1}{f_2} = 10{f_2} + 10{f_1} - 20

\Rightarrow 10{f_1} + 10{f_2} - {f_1}{f_2} = + 20

For f1 = 18 cm and f2 = 20 cm, the above equation satisfies.

Q120

If the refractive index of the material of a prism is

\cot \left(\dfrac{A}{2}\right)

, where

A

is the angle of prism then the angle of minimum deviation will be

A

\pi-2 \mathrm{~A}

B

\dfrac{\pi}{2}-2 \mathrm{~A}

C

\pi-\mathrm{A}

D

\dfrac{\pi}{2}-\mathrm{A}

Correct Answer

Option A

Solution

\begin{aligned} & \cot \frac{\mathrm{A}}{2}=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right)}{\sin \frac{\mathrm{A}}{2}} \\ & \Rightarrow \cos \frac{\mathrm{A}}{2}=\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right) \\ & \frac{\mathrm{A}+\delta_{\min }}{2}=\frac{\pi}{2}-\frac{\mathrm{A}}{2} \\ & \delta_{\min }=\pi-2 \mathrm{~A} \end{aligned}%