Geometrical Optics — Page 7 | JEE Physics

Q61

Light enters from air into a given medium at an angle of

45^{\circ}

with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of

15^{\circ}

from its original direction. The refractive index of the medium is:

A 1.732

B 1.333

C 1.414

D 2.732

Correct Answer

Option C

Solution

Let, refractive index of medium = $\mu$ $\therefore$

r + 15^\circ = 45^\circ

\Rightarrow r = 30^\circ

Using Snell's law,

1\,.\,\sin 45^\circ = \sin 30^\circ \times \mu

\Rightarrow {1 \over {\sqrt 2 }} = {1 \over 2} \times \mu

\Rightarrow \mu = {2 \over {\sqrt 2 }} = \sqrt 2 = 1.414

Q62

Two objects A and B are placed at 15 cm and 25 cm from the pole in front of a concave mirror having radius of curvature 40 cm. The distance between images formed by the mirror is _______________.

A 60 cm

B 40 cm

C 160 cm

D 100 cm

Correct Answer

Option C

Solution

Using Mirror formula

\begin{aligned} & \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\\\ & \frac{1}{v}=\frac{1}{f}-\frac{1}{f} \\\\ & v=\frac{u f}{u-f} \end{aligned}

For object $\mathrm{A}, \mathrm{u}_{\mathrm{i}}=-15 \mathrm{~cm}, \mathrm{f}=-20 \mathrm{~cm}, \mathrm{~v}_1=$ ?

\begin{aligned} & \mathrm{v}_1=\frac{\mathrm{u}_1 \mathrm{f}}{\mathrm{u}_1-\mathrm{f}}=\frac{(-15)(-20)}{(-15)-(20)}=\frac{+300}{5} \\\\ & \mathrm{v}_1=+60 \mathrm{~cm}(+ ve, virtual) \end{aligned}

For object $\mathrm{B}, \mathrm{u}_2=-25 \mathrm{~cm}, \mathrm{f}=-20 \mathrm{~cm} \mathrm{v}_2=$ ?

\begin{aligned} & \mathrm{v}_2=\frac{\mathrm{u}_2 \mathrm{f}}{\mathrm{u}_2-\mathrm{f}}=\frac{(-25)(-20)}{(-25)-(-20)}=\frac{500}{-5} \\\\ & \mathrm{v}_2=-100 \mathrm{~cm}(- ve, real) \end{aligned}

Hence, the distance between images formed by the mirror is, d = 60 – (–100) = 160 cm

Q63

A microscope is focused on an object at the bottom of a bucket. If liquid with refractive index

\dfrac{5}{3}

is poured inside the bucket, then the microscope has to be raised by

30 \mathrm{~cm}

to focus the object again. The height of the liquid in the bucket is :

A

50 \mathrm{~cm}

B

18 \mathrm{~cm}

C

75 \mathrm{~cm}

D

12 \mathrm{~cm}

Correct Answer

Option C

Solution

Shift $=\left(d-\dfrac{d}{\mu}\right)=30 \mathrm{~cm}$

\begin{aligned} & \Rightarrow d\left[1-\frac{1}{\frac{5}{3}}\right]=30 \\\\ & \Rightarrow d=\frac{30 \times 5}{2}=75 \mathrm{~cm} \end{aligned}

Q64

A thin prism

P_1

with an angle

6^{\circ}

and made of glass of refractive index

1.54

is combined with another prism

P_2

made from glass of refractive index

1.72

to produce dispersion without average deviation. The angle of prism

P_2

is

A

4.5^{\circ}

B

7.8^{\circ}

C

1.3^{\circ}

D

6^{\circ}

Correct Answer

Option A

Solution

({\mu _1} - 1){A_1} = ({\mu _2} - 1){A_2}

\Rightarrow (1.54 - 1)6 = (1.72 - 1){A_2}

{A_2} = \left( {{{0.54} \over {0.72}} \times 6} \right) = {{18} \over 4} = \left( {{9 \over 2}} \right) = 4.5^\circ

Q65

When a beam of white light is allowed to pass through convex lens parallel to principal axis, the different colours of light converge at different point on the principle axis after refraction. This is called :

A Spherical aberration

B Scattering

C Polarisation

D Chromatic aberration

Correct Answer

Option D

Solution

The phenomenon described in the statement is called chromatic aberration.

Chromatic aberration is a type of optical aberration that occurs when a lens is unable to focus different colors of light at the same point after refraction.

This causes the different colors of light to converge at different points on the principal axis, resulting in blurred or distorted images.

In the case of a convex lens, the refractive index of the lens is different for different colors of light.

The blue light, having the shortest wavelength, refracts the most and the red light, having the longest wavelength, refracts the least.

This causes the blue light to converge at a point closer to the lens than the red light, resulting in a blurred image.

Spherical aberration, on the other hand, is a type of optical aberration that occurs when a spherical lens is unable to focus all the incident light rays at a single point, resulting in a blurred image.

Polarization refers to the orientation of the electric field of light waves, and scattering refers to the phenomenon of light being redirected in different directions due to interaction with matter.

Q66

The refracting angle of a prism is A and refractive index of the material of the prism is cot (A/2). Then the angle of minimum deviation will be -

A 180

-

2A

B 90

-

A

C 180 + 2A

D 180

-

3A

Correct Answer

Option A

Solution

\mu = {{\sin \left( {{{{\delta _m} + A} \over 2}} \right)} \over {\sin (A/2)}} = \cot A/2

\Rightarrow \cos A/2 = \sin \left( {{{{\delta _m} + A} \over 2}} \right)

\Rightarrow {\pi \over 2} - {A \over 2} = {{{\delta _m} + A} \over 2}

\Rightarrow \pi - 2A = {\delta _m}

Q67

A vessel of depth '

d

' is half filled with oil of refractive index

n_{1}

and the other half is filled with water of refractive index

n_{2}

. The apparent depth of this vessel when viewed from above will be-

A

\dfrac{2 d\left(n_{1}+n_{2}\right)}{n_{1} n_{2}}

B

\dfrac{d\left(n_{1}+n_{2}\right)}{2 n_{1} n_{2}}

C

\dfrac{d n_{1} n_{2}}{2\left(n_{1}+n_{2}\right)}

D

\dfrac{d n_{1} n_{2}}{\left(n_{1}+n_{2}\right)}

Correct Answer

Option B

Solution

To find the apparent depth of the vessel when viewed from above, we can calculate the apparent depths of the oil and water separately and then add them together.

The formula to find the apparent depth (

h_{apparent}

) is:

h_{apparent} = \frac{h_{real}}{n}

Where

h_{real}

is the actual depth, and

n

is the refractive index of the medium. For the oil (with depth

\frac{d}{2}

and refractive index

n_1

):

h_{oil} = \frac{\frac{d}{2}}{n_{1}}

For the water (with depth

\frac{d}{2}

and refractive index

n_2

):

h_{water} = \frac{\frac{d}{2}}{n_{2}}

Now, add the two apparent depths together to find the total apparent depth:

h_{total} = h_{oil} + h_{water} = \frac{\frac{d}{2}}{n_{1}} + \frac{\frac{d}{2}}{n_{2}}

Combine the terms:

h_{total} = \frac{d}{2}\left(\frac{1}{n_{1}} + \frac{1}{n_{2}}\right)

Now, find a common denominator for the fractions:

h_{total} = \frac{d}{2}\left(\frac{n_{1} + n_{2}}{n_{1}n_{2}}\right)

Multiply the fractions:

h_{total} = \frac{d(n_{1} + n_{2})}{2n_{1}n_{2}}

The correct answer is:

\frac{d\left(n_{1}+n_{2}\right)}{2 n_{1} n_{2}}

Q68

An ice cube has a bubble inside. When viewed from one side the apparent distance of the bubble is

12 \mathrm{~cm}

. When viewed from the opposite side, the apparent distance of the bubble is observed as

4 \mathrm{~cm}

. If the side of the ice cube is

24 \mathrm{~cm}

, the refractive index of the ice cube is

A

\dfrac{3}{2}

B

\dfrac{4}{3}

C

\dfrac{2}{3}

D

\dfrac{6}{5}

Correct Answer

Option A

Solution

Let's denote the true distance of the bubble from one side of the ice cube as

x

and the refractive index of the ice cube as

n

.

We will use the formula for apparent depth, which states that the ratio of the true depth to the apparent depth is equal to the refractive index:

n = \frac{\text{True depth}}{\text{Apparent depth}}

When viewing the bubble from one side, the true depth is

x

and the apparent depth is

12 \mathrm{~cm}

. Using the formula:

n = \frac{x}{12}

When viewing the bubble from the opposite side, the true depth is

24 - x

(since the side of the ice cube is

24 \mathrm{~cm}

) and the apparent depth is

4 \mathrm{~cm}

. Using the formula:

n = \frac{24 - x}{4}

Now we have a system of two equations with two variables: 1)

n = \frac{x}{12}

2)

n = \frac{24 - x}{4}

We can solve this system by setting the two expressions for

n

equal to each other:

\frac{x}{12} = \frac{24 - x}{4}

To solve for

x

, first multiply both sides by

12

:

x = 3(24 - x)

x = 72 - 3x

Add

3x

to both sides:

4x = 72

Divide by

4

:

x = 18

Now that we have the value of

x

, we can find the refractive index

n

using either equation 1 or 2. Using equation 1:

n = \frac{18}{12} = \frac{3}{2}

Therefore, the refractive index of the ice cube is

\frac{3}{2}

.

Q69

When one light ray is reflected from a plane mirror with

30^{\circ}

angle of reflection, the angle of deviation of the ray after reflection is :

A

140^{\circ}

B

130^{\circ}

C

120^{\circ}

D

110^{\circ}

Correct Answer

Option C

Solution

When a light ray is reflected from a plane mirror, the angle of incidence (i) is equal to the angle of reflection (r).

In this case, the angle of reflection is given as

30^{\circ}

, so the angle of incidence is also

30^{\circ}

.

The angle of deviation (D) is the angle between the incident ray and the reflected ray.

To find this angle, consider the fact that the angle between the incident ray and the normal to the mirror and the angle between the reflected ray and the normal add up to

180^{\circ}

, since they are supplementary angles. Thus, we have:

i + r + D = 180^{\circ}

Since

i = r

, we can rewrite the equation as:

2i + D = 180^{\circ}

Substitute the value of the angle of incidence:

2(30^{\circ}) + D = 180^{\circ}

60^{\circ} + D = 180^{\circ}

Solve for the angle of deviation (D):

D = 180^{\circ} - 60^{\circ} = 120^{\circ}

So, the angle of deviation of the ray after reflection is

120^{\circ}

.

Q70

A 2 meter long scale with least count of

0.2 \mathrm{~cm}

is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at

80 \mathrm{~cm}

mark and

1 \mathrm{~m}

mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at

180 \mathrm{~cm}

mark. The

\%

error in the estimation of focal length is:

A 1.70

B 0.51

C 1.02

D 0.85

Correct Answer

Option A

Solution

In this problem, you are asked to find the percentage error in the estimation of the focal length of a convex lens using a 2-meter long scale with a least count of 0.2 cm.

First, let's determine the object distance (u), image distance (v), and focal length (f) of the lens.

Object distance (u): It's the distance between the object pin and the convex lens.

The object pin is at the 80 cm mark, and the convex lens is at the 1 m (100 cm) mark, so the object distance is

u = 100 - 80 = 20~cm

.

Image distance (v): It's the distance between the image pin and the convex lens.

The image pin is at the 180 cm mark, and the convex lens is at the 1 m (100 cm) mark, so the image distance is

v = 180 - 100 = 80~cm

. Focal length (f): Using the lens formula, we can calculate the focal length:

\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{80} + \frac{1}{20} = \frac{5}{80}

So,

f = \frac{80}{5} = 16~cm

.

Now, we will calculate the error in the focal length (df) using the given least count (0.2 cm).

The error in the object distance and image distance will both be 0.2 cm.

Error in the focal length (df): We can use the formula for the error in the focal length:

\frac{df}{f^2} = \frac{0.2 \times 2}{6400} + \frac{0.2 \times 2}{400}

Solving for df:

df = \frac{16 \times 16 \times 0.2 \times 6800 \times 2}{6400 \times 400} = 0.136 \times 2

Percentage error in the focal length: Finally, we will calculate the percentage error using the formula:

\frac{df}{f} = \frac{0.0085 \times 2}{1} = 1.70

So, the percentage error in the estimation of the focal length is 1.70%.