Geometrical Optics — Page 8 | JEE Physics

Q71

The critical angle for a denser-rarer interface is

45^{\circ}

. The speed of light in rarer medium is

3 \times 10^{8} \mathrm{~m} / \mathrm{s}

. The speed of light in the denser medium is:

A

2 .12 \times 10^{8} \mathrm{~m} / \mathrm{s}

B

5 \times 10^{7} \mathrm{~m} / \mathrm{s}

C

\sqrt{2} \times 10^{8} \mathrm{~m} / \mathrm{s}

D

3.12 \times 10^{7} \mathrm{~m} / \mathrm{s}

Correct Answer

Option A

Solution

To find the speed of light in the denser medium, we can use Snell's Law at the critical angle, where the angle of refraction is

90^{\circ}

. Snell's Law states:

n_1 \sin{\theta_1} = n_2 \sin{\theta_2}

where

n_1

and

n_2

are the indices of refraction for the denser and rarer media, respectively, and

\theta_1

and

\theta_2

are the angles of incidence and refraction, respectively. In this case, we have:

n_1 \sin{45^{\circ}} = n_2 \sin{90^{\circ}}

The critical angle is given as

45^{\circ}

, and the speed of light in the rarer medium is given as

3 \times 10^8 \mathrm{~m/s}

. We can find the index of refraction of the rarer medium using the formula:

n = \frac{c}{v}

where

c

is the speed of light in a vacuum (

3 \times 10^8 \mathrm{~m/s}

), and

v

is the speed of light in the medium. For the rarer medium, we have:

n_2 = \frac{3 \times 10^8 \mathrm{~m/s}}{3 \times 10^8 \mathrm{~m/s}} = 1

Now we can rewrite Snell's Law as:

n_1 \sin{45^{\circ}} = 1 \cdot 1

n_1 \sin{45^{\circ}} = 1

n_1 = \frac{1}{\sin{45^{\circ}}}

Since

\sin{45^{\circ}} = \frac{1}{\sqrt{2}}

, we have:

n_1 = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}

Now, we can find the speed of light in the denser medium using the formula for the index of refraction:

v_1 = \frac{c}{n_1}

v_1 = \frac{3 \times 10^8 \mathrm{~m/s}}{\sqrt{2}}

v_1 = 2.12 \times 10^8 \mathrm{~m/s}

So, the speed of light in the denser medium is

2.12 \times 10^8 \mathrm{~m/s}

.

Q72

An object is placed at a distance of 12 cm in front of a plane mirror. The virtual and erect image is formed by the mirror. Now the mirror is moved by 4 cm towards the stationary object. The distance by which the position of image would be shifted, will be

A 4 cm towards mirror

B 2 cm towards mirror

C 8 cm away from mirror

D 8 cm towards mirror

Correct Answer

Option D

Solution

$\therefore$ Shifting of image will be 8 cm towards mirror.

Q73

In a reflecting telescope, a secondary mirror is used to:

A make chromatic aberration zero

B remove spherical aberration

C reduce the problem of mechanical support

D move the eyepiece outside the telescopic tube

Correct Answer

Option D

Solution

Reflecting telescopes, also known as reflectors, use a set of mirrors instead of lenses to gather and focus light.

The primary mirror (usually a concave mirror) gathers the incoming light and reflects it to a focus point.

The secondary mirror is used to redirect this focused light out to where it can be conveniently observed.

In other words, the secondary mirror in a reflecting telescope is used to move the eyepiece outside the telescopic tube, where the image can be comfortably viewed.

This is especially important for large telescopes, where the focus point may be inside the telescope tube and inaccessible without a secondary mirror.

Q74

A monochromatic light wave with wavelength

\lambda_{1}

and frequency

v_{1}

in air enters another medium. If the angle of incidence and angle of refraction at the interface are

45^{\circ}

and

30^{\circ}

respectively, then the wavelength

\lambda_{2}

and frequency

v_{2}

of the refracted wave are:

A

\lambda_{2}=\lambda_{1}, v_{2}=\dfrac{1}{\sqrt{2}} v_{1}

B

\lambda_{2}=\lambda_{1}, v_{2}=\sqrt{2} v_{1}

C

\lambda_{2}=\sqrt{2} \lambda_{1}, v_{2}=v_{1}

D

\lambda_{2}=\dfrac{1}{\sqrt{2}} \lambda_{1}, v_{2}=v_{1}

Correct Answer

Option D

Solution

When a light wave moves from one medium to another, its speed and wavelength may change, but its frequency remains constant because it is determined by the source of the light.

This is because frequency depends on the oscillations of the source, which do not change when entering a different medium.

Snell's law, which describes the relationship between the angles of incidence and refraction, can be written as: $n_1 \sin(\theta_1) = n_2 \sin(\theta_2),$ where $n_1$ and $n_2$ are the refractive indices of the two media, and $\theta_1$ and $\theta_2$ are the angles of incidence and refraction, respectively.

The refractive index of a medium is also related to the speed of light in that medium: $n = \dfrac{c}{v},$ where $c$ is the speed of light in a vacuum, and $v$ is the speed of light in the medium.

From the above relations, we see that as light enters a medium with a higher refractive index (and hence a lower speed), its wavelength decreases.

The frequency remains the same.

Given that the angle of incidence is $45^\circ$ and the angle of refraction is $30^\circ$ , the ratio of the refractive indices is: $\dfrac{n_2}{n_1} = \dfrac{\sin(\theta_1)}{\sin(\theta_2)} = \dfrac{\sin(45^\circ)}{\sin(30^\circ)} = \sqrt{2}.$ This indicates that the speed of light (and hence the wavelength) decreases by a factor of $\sqrt{2}$ as it enters the second medium.

The frequency remains the same.

Therefore, the correct answer is $\lambda_2 = \dfrac{1}{\sqrt{2}} \lambda_1, \quad v_2 = v_1.$

Q75

Identify the physical quantity that cannot be measured using spherometer :

A Radius of curvature of concave surface

B Specific rotation of liquids

C Thickness of thin plates

D Radius of curvature of convex surface

Correct Answer

Option B

Solution

Spherometer can be used to measure curvature of surface.

Q76

The refractive index of a prism with apex angle

A

is

\cot A / 2

. The angle of minimum deviation is :

A

\delta_m=180^{\circ}-3 \mathrm{~A}

B

\delta_m=180^{\circ}-4 A

C

\delta_m=180^{\circ}-2 A

D

\delta_m=180^{\circ}-A

Correct Answer

Option C

Solution

\begin{aligned} & \mu=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\ & \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\ & \sin \left(\frac{\pi}{2}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_m}{2}\right) \\ & \frac{\pi}{2}-\frac{A}{2}=\frac{A}{2}+\frac{\delta m}{2} \\ & \delta_m=\pi-2 A \end{aligned}

Q77

A convex mirror of radius of curvature

30 \mathrm{~cm}

forms an image that is half the size of the object. The object distance is :

A

-

45 cm

B

-

15 cm

C 45 cm

D 15 cm

Correct Answer

Option B

Solution

Given

\mathrm{R}=30 \mathrm{~cm}

\mathrm{f}=\mathrm{R} / 2=+15 \mathrm{~cm}

Magnification

(\mathrm{m})= \pm \frac{1}{2}

For convex mirror, virtual image is formed for real object. Therefore,

\mathrm{m}

is +ve

\begin{aligned} & \frac{1}{2}=\frac{f}{f-u} \\ & u=-15 \mathrm{~cm} \end{aligned}

Q78

A biconvex lens of refractive index 1.5 has a focal length of

20 \mathrm{~cm}

in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:

A

-

16 cm

B +16 cm

C +160 cm

D

-

160 cm

Correct Answer

Option D

Solution

\begin{aligned} & \mu_1=1.5 \\ & \mu_m=1.6 \\ & f_a=20 \mathrm{~cm} \\ & \text{ As } \frac{f_m}{f_a}=\frac{\left(\mu_1-1\right) \mu_m}{\left(\mu_1-\mu_m\right)} \\ & \frac{f_m}{20}=\frac{(1.5-1) 1.6}{(1.5-1.6)} \\ & f_m=-160 \mathrm{~cm} \end{aligned}

Q79

If the distance between object and its two times magnified virtual image produced by a curved mirror is

15 \mathrm{~cm}

, the focal length of the mirror must be:

A

-10

cm

B

-12

cm

C 15 cm

D 10/3 cm

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{m}=2=\frac{-v}{u} \\ & 2=\frac{-(15-u)}{-u} \\ & 2 u=15-u \\ & 3 u=15 \Rightarrow u=5 \mathrm{~cm} \\ & v=15-u=15-5=10 \mathrm{~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 \mathrm{~cm} \end{aligned}

Q80

Given below are two statements : Statement (I) : When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side. Statement (II) : Concave lens always forms a virtual and erect image. In the light of the above statements, choose the correct answer from the options given below :

A Statement I is false but Statement II is true

B Both Statement I and Statement II are false

C Both Statement I and Statement II are true

D Statement I is true but Statement II is false

Correct Answer

Option A

Solution

Let's analyze each statement carefully: Statement (I): When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side.

To understand this statement, let's recall the nature of images formed by concave lenses.

A concave lens (diverging lens) always forms images that are virtual, erect, and smaller than the object, regardless of the object's position.

The term "centre of curvature" is typically associated with mirrors rather than lenses.

Therefore, this statement is incorrect because a concave lens does not form a real image that could be said to be at the centre of curvature on the other side.

Statement (II): Concave lens always forms a virtual and erect image.

Now, considering the second statement, a concave lens (diverging lens) indeed always forms virtual, erect, and diminished images.

This is a fundamental property of concave lenses.

Given this analysis: The correct answer is Option A: Statement I is false but Statement II is true.