Geometrical Optics — Page 9 | JEE Physics

Q81

An effective power of a combination of 5 identical convex lenses which are kept in contact along the principal axis is

25 \mathrm{D}

. Focal length of each of the convex lens is:

A 50 cm

B 20 cm

C 25 cm

D 500 cm

Correct Answer

Option B

Solution

When we have a combination of identical lenses in contact, the effective power (

P_{\text{eff}}

) of the combination can be calculated as the sum of the powers of all the individual lenses.

This is because the lenses are in direct contact, and their powers effectively add up.

Given that the effective power of a combination of 5 identical convex lenses is

25 \mathrm{D}

, we can use the formula for the effective power of the combination:

P_{\text{eff}} = nP

where: $P_{\text{eff}}$ is the effective power of the combination, $n$ is the number of lenses, and $P$ is the power of each individual lens.

Given $n = 5$ and $P_{\text{eff}} = 25 \mathrm{D}$ , we can solve for $P$ , the power of each lens:

25 \mathrm{D} = 5P

Dividing both sides by 5:

P = \frac{25 \mathrm{D}}{5} = 5 \mathrm{D}

The power of a lens ( $P$ ) is related to its focal length ( $f$ ) by the equation:

P = \frac{1}{f}

where $P$ is in diopters (D) and $f$ is in meters. Thus:

5 \mathrm{D} = \frac{1}{f}

Solving for $f$ gives:

f = \frac{1}{5} \text{ meters} = \frac{1}{5} \times 100 \text{ cm} = 20 \text{ cm}

Therefore, the focal length of each of the convex lenses is 20 cm. So, the correct answer is Option B: 20 cm.

Q82

In an experiment to measure focal length (

f

) of convex lens, the least counts of the measuring scales for the position of object (u) and for the position of image (v) are

\Delta u

and

\Delta v

, respectively. The error in the measurement of the focal length of the convex lens will be:

A

2 f\left[\dfrac{\Delta \mathrm{u}}{\mathrm{u}}+\dfrac{\Delta \mathrm{v}}{\mathrm{v}}\right]

B

f\left[\dfrac{\Delta \mathrm{u}}{\mathrm{u}}+\dfrac{\Delta \mathrm{v}}{\mathrm{v}}\right]

C

f^2\left[\dfrac{\Delta \mathrm{u}}{\mathrm{u}^2}+\dfrac{\Delta \mathrm{v}}{\mathrm{v}^2}\right]

D

\dfrac{\Delta \mathrm{u}}{\mathrm{u}}+\dfrac{\Delta \mathrm{v}}{\mathrm{v}}

Correct Answer

Option C

Solution

First, let's understand the relationship between the object distance (

u

), the image distance (

v

), and the focal length (

f

) of a convex lens, which is given by the lens formula:

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

Now, to find the error in the focal length (

\Delta f

) due to the errors in the measurements of

u

and

v

(

\Delta u

and

\Delta v

, respectively), we have to differentiate the lens formula with respect to

u

and

v

, keeping in mind the propagation of error. By differentiating both sides of the lens formula with respect to

v

and

u

, and also considering the negative reciprocal relation (given

f

is a constant for a specific lens), we have:

\frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2}

Rearranging this equation to find

\Delta f

, we get:

\Delta f = f^2 \left( \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} \right)

Therefore, the correct option showing the error in the measurement of the focal length of the convex lens, taking into account the least counts (

\Delta u

and

\Delta v

) of the measuring scales for the position of the object (

u

) and for the position of the image (

v

), is: Option C:

f^2\left[\frac{\Delta \mathrm{u}}{\mathrm{u}^2}+\frac{\Delta \mathrm{v}}{\mathrm{v}^2}\right]

Q83

Critical angle of incidence for a pair of optical media is

45^{\circ}

. The refractive indices of first and second media are in the ratio:

A

1: 2

B

1: \sqrt{2}

C

2: 1

D

\sqrt{2}: 1

Correct Answer

Option D

Solution

The critical angle is the angle of incidence at which light is refracted along the boundary, meaning the angle of refraction is

90^{\circ}

.

The relationship between the critical angle and the refractive indices of two media can be understood using Snell's Law, given as:

n_1 \sin(\theta_c) = n_2 \sin(90^{\circ})

Here,

n_1

is the refractive index of the first medium,

n_2

is the refractive index of the second medium, and

\theta_c

is the critical angle. Since

\sin(90^{\circ}) = 1

, the equation simplifies to:

n_1 \sin(\theta_c) = n_2

Given that the critical angle

\theta_c

is

45^{\circ}

, we have:

\sin(45^{\circ}) = \frac{\sqrt{2}}{2}

Substituting this value in the simplified Snell's Law equation, we get:

n_1 \left( \frac{\sqrt{2}}{2} \right) = n_2

Therefore,

n_1 = n_2 \cdot \frac{2}{\sqrt{2}}

n_1 = n_2 \cdot \sqrt{2}

Hence, the ratio of the refractive indices of the first and second media is:

n_1 : n_2 = \sqrt{2} : 1

The correct answer is therefore: Option D

\sqrt{2}: 1

Q84

A convex lens made of glass (refractive index = 1.5) has focal length 24 cm in air. When it is totally immersed in water (refractive index = 1.33), its focal length changes to

A 96 cm

B 72 cm

C 24 cm

D 48 cm

Correct Answer

Option A

Solution

Using lens formula, For Ist case :

{1 \over f} = \left( {{a^{{\mu _g}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

where,

{R_1},{R_2}

= radii of refracting surfaces $f$ = focal length

{a^{{\mu _g}}} = {{^{{\mu _g}}} \over {^{{\mu _a}}}} = {{1.5} \over 1} = 1.5

So,

{1 \over {24}} = (1.5 - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

(As f = 24 cm given)

\Rightarrow {1 \over {24}} = 0.5\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

\Rightarrow {1 \over {12}} = \left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

\Rightarrow {1 \over {{R_1}}} - {1 \over {{R_2}}} = {1 \over {12}}

.... (i) For IInd case:

{1 \over f} = \left( {w{\mu _g} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

\Rightarrow {1 \over f} = \left( {{{1.5} \over {1.33}} - 1} \right)\left( {{1 \over {12}}} \right)

..... from (i)

\Rightarrow {1 \over f} = {{17} \over {133}} \times {1 \over {12}} \Rightarrow f = {{133 \times 12} \over {17}}

\Rightarrow f = 93.88

cm $\approx$ 94 cm Hence, option 1 is correct.

Q85

Given is a thin convex lens of glass (refractive index

\mu

) and each side having radius of curvature

R

. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that the image gets formed on the object itself?

A

R / \mu

B

R /(2 \mu-1)

C

\mathrm{R} /(2 \mu-3)

D

\mu \mathrm{R}

Correct Answer

Option B

Solution

We know, for a combination of a lens and a mirror,

{P_{comb}} = 2{P_l} + {P_m}

...... (1) Also using lens maker formula,

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

So using eq. (i),

- {1 \over {{f_{comb}}}} = 2\left( {{1 \over {{f_l}}}} \right) - {1 \over {{f_m}}}

\Rightarrow - {1 \over {{f_{comb}}}} = {{4(\mu - 1)} \over R} - {1 \over {\left( { - {R \over 2}} \right)}}

\Rightarrow - {1 \over {{f_{comb}}}} = {{4(\mu - 1)} \over R} + {2 \over R}

\Rightarrow - {1 \over {{f_{comb}}}} = {2 \over R}(2\mu - 2 + 1) = {2 \over R}(2\mu - 1)

\Rightarrow {f_{comb}} = {{ - R} \over {2(2\mu - 1)}}

\Rightarrow {R_{comb}} = 2{f_{comb}} = {{ - 2R} \over { - 2(2\mu - 1)}} = {{ - R} \over {2\mu - 1}}

Hence, distance

= {R \over {2\mu - 1}}

Q86

In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A . If an object placed at 13 cm from the vertex of the meniscus in A forms an image with a magnification of

{ }^{\prime}-2^{\prime}

then the radius of curvature of meniscus is :

A

\dfrac{2}{3} \text{ cm}

B

\dfrac{4}{3} \text{ cm}

C

\dfrac{1}{3} \text{ cm}

D 1 cm

Correct Answer

Option A

Solution

\begin{aligned} & \frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}} \frac{\mathrm{n}_2-\mathrm{n}_1}{\mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}-\frac{1.3}{-13}=\frac{0.1}{\mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}=\frac{1-\mathrm{R}}{10 \mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}=\frac{1-\mathrm{R}}{10 \mathrm{R}} \\ & \mathrm{~m}=\frac{\mathrm{v} / \mathrm{n}_2}{\mathrm{u} / \mathrm{n}_1} \\ & -2 \times \frac{(-13)}{1.3}=\frac{10 \mathrm{R}}{1-\mathrm{R}} \\ & \mathrm{R}=\frac{2}{3} \mathrm{~cm} \end{aligned}

Q87

What is the lateral shift of a ray refracted through a parallel-sided glass slab of thickness '

h

' in terms of the angle of incidence '

i

' and angle of refraction '

r

', if the glass slab is placed in air medium?

A

\mathrm{h}

B

\dfrac{h \cos (i-r)}{\sin r}

C

\dfrac{\mathrm{h} \tan (\mathrm{i}-\mathrm{r})}{\tan \mathrm{r}}

D

\dfrac{h \sin (i-r)}{\cos r}

Correct Answer

Option D

Solution

We can derive the lateral shift (d) by considering the geometry of a ray entering a parallel-sided slab.

Here’s a step‐by‐step explanation: The slab has a thickness

h

(measured perpendicular to the surfaces). When the ray enters the glass from air at angle

i

(with respect to the normal), it refracts inside and makes an angle

r

. Inside the slab, the ray travels a distance

L = \frac{h}{\cos r}

because the vertical component of the path must cover a distance

h

. The ray inside the slab is shifted laterally (parallel to the surface). Its lateral displacement over the distance

L

is

L \sin r = \frac{h \sin r}{\cos r}.

However, the emergent ray is parallel to the incident ray but not collinear.

By constructing the appropriate geometry (extending the emergent ray backwards until it meets the incident ray's extension at the top interface), you find that the lateral shift between the incident ray and the emergent ray is given by

d = \frac{h \sin(i - r)}{\cos r}.

This expression correctly accounts for both the refraction inside the slab and the geometry of the displacement.

Thus, the correct answer is: Option D:

\frac{h \sin (i-r)}{\cos r}.

Q88

The refractive index of the material of a glass prism is

\sqrt{3}

. The angle of minimum deviation is equal to the angle of the prism. What is the angle of the prism?

A

60^{\circ}

B

50^{\circ}

C

58^{\circ}

D

48^{\circ}

Correct Answer

Option A

Solution

\textbf{Step 1: Symmetry at Minimum Deviation}

In a prism, when the deviation is minimum, the path of light is symmetric.

This means that the angle of incidence ( $i$ ) is equal to the angle of emergence ( $i'$ ), and the light inside the prism makes equal angles with the prism faces.

If the prism angle is

A

, then the refracted angle at each interface is:

r = \frac{A}{2}

\textbf{Step 2: Relating Deviation to the Angles}

The formula for the deviation ( $D$ ) in a prism is given by:

D = i + i' - A

At minimum deviation,

i = i'

, so:

D_{\text{min}} = 2i - A

The special condition given is that the minimum deviation is equal to the prism angle:

D_{\text{min}} = A

Thus:

A = 2i - A \quad \Longrightarrow \quad 2i = 2A \quad \Longrightarrow \quad i = A

\textbf{Step 3: Applying Snell's Law}

At the first surface, Snell's law gives:

\sin i = \mu \sin r

Substitute the values

i = A

and

r = \frac{A}{2}

:

\sin A = \mu \sin\frac{A}{2}

Given that the refractive index

\mu = \sqrt{3}

, we have:

\sin A = \sqrt{3} \sin\frac{A}{2}

\textbf{Step 4: Solving the Equation}

Recall the double-angle formula for sine:

\sin A = 2 \sin\frac{A}{2} \cos\frac{A}{2}

Substitute this into the previous equation:

2 \sin\frac{A}{2} \cos\frac{A}{2} = \sqrt{3} \sin\frac{A}{2}

Assuming

\sin\frac{A}{2} \neq 0

, we can divide both sides by

\sin\frac{A}{2}

:

2 \cos\frac{A}{2} = \sqrt{3}

Solve for

\cos\frac{A}{2}

:

\cos\frac{A}{2} = \frac{\sqrt{3}}{2}

Since:

\cos 30^\circ = \frac{\sqrt{3}}{2}

It follows:

\frac{A}{2} = 30^\circ \quad \Longrightarrow \quad A = 60^\circ

\textbf{Final Answer:}

The angle of the prism is

60^\circ

.

Q89

What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from 2.5D ? ['D' stands for dioptre]

A 0.04

B 0.1

C 0.40

D 0.01

Correct Answer

Option A

Solution

P_1 = 2.5\,D,\quad f_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4\,m

After an increase in optical power by 0.1 D:

P_2 = 2.5\,D + 0.1\,D = 2.6\,D,\quad f_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846\,m

The change in focal length is:

\Delta f = f_1 - f_2 = 0.4\,m - 0.3846\,m \approx 0.0154\,m

The relative decrease in focal length is then calculated as:

\text{Relative decrease} = \frac{\Delta f}{f_1} = \frac{0.0154}{0.4} \approx 0.0385 \approx 0.04

Thus, the relative decrease in focal length is approximately 0.04.

Q90

A thin plano convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2. When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length 0.2 m . The radius of curvature of the curved surface of the lens is

A 0.15 m

B 0.10 m

C 0.25 m

D 0.20 m

Correct Answer

Option B

Solution

\frac{1}{f} = \frac{2(n_2 - n_1)}{n_1 R}

For a thin plano‑convex lens made of glass (refractive index

n_2 = 1.5

) immersed in a liquid (refractive index

n_1 = 1.2

), when the plane side is silver‐coated, a ray entering the curved surface from the liquid is refracted into the glass, reflects off the coated plane (thereby reversing its direction) and finally refracts back into the liquid at the same curved surface.

In the paraxial approximation the net effect is equivalent to a mirror in the liquid with an effective focal length

f

. A ray that is incident parallel to the optical axis (at height

h

) will, after this combined refraction–reflection–refraction process, emerge with an angular deviation corresponding to a mirror having a power given by:

\frac{h}{f} = \frac{2(n_2 - n_1)}{n_1R}\, h\,.

This gives the relation:

\frac{1}{f} = \frac{2(n_2 - n_1)}{n_1 R}\,.

Given that the effective focal length is

f = 0.2\,\text{m}

, substitute the values:

\frac{1}{0.2} = \frac{2(1.5 - 1.2)}{1.2\,R}\,.

Simplify the numerator:

\frac{1}{0.2} = \frac{2(0.3)}{1.2\,R} = \frac{0.6}{1.2\,R}\,.

Since

\frac{0.6}{1.2} = 0.5,

the equation becomes:

5 = \frac{0.5}{R}\,.

Solving for

R

:

R = \frac{0.5}{5} = 0.10\,\text{m}\,.

Thus, the radius of curvature of the curved surface of the lens is

0.10\,\text{m}

.