A light ray is incident on a transparent sphe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $1$. Applying Snell's Law at the first point of incidence: $1 \cdot \sin(45^{\circ}) = \sqrt{2} \cdot \sin(r) \Rightarrow \sin(r) = \frac{1}{\sqrt

Vedclass · Accepted Answer

$270$

Vedclass · Answer

(A) $1$. Applying Snell's Law at the first point of incidence: $1 \cdot \sin(45^{\circ}) = \sqrt{2} \cdot \sin(r) \Rightarrow \sin(r) = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = 0.5 \Rightarrow r = 30^{\circ}$.$2$. Deviation at the first refraction: $\delta_1 = i - r = 45^{\circ} - 30^{\circ} = 15^{\circ}$.$3$. At each of the two internal reflections,the angle of incidence is $r = 30^{\circ}$. The deviation at each reflection is $\delta = 180^{\circ} - 2i_{reflection} = 180^{\circ} - 2(30^{\circ}) = 120^{\circ}$.$4$. For two internal reflections,total deviation $\delta_2 + \delta_3 = 120^{\circ} + 120^{\circ} = 240^{\circ}$.$5$. At the final refraction,the angle of incidence is $r = 30^{\circ}$ and the angle of emergence is $e = 45^{\circ}$. The deviation is $\delta_4 = e - r = 45^{\circ} - 30^{\circ} = 15^{\circ}$.$6$. Total deviation $\delta_{total} = \delta_1 + \delta_2 + \delta_3 + \delta_4 = 15^{\circ} + 120^{\circ} + 120^{\circ} + 15^{\circ} = 270^{\circ}$.

$A$ light ray is incident on a transparent sphere of refractive index $\mu = \sqrt{2}$ at an angle of incidence $i = 45^{\circ}$. What is the total deviation of a ray that enters the sphere,undergoes two internal reflections,and then refracts out into the air (in $^{\circ}$)?

Similar Questions

Let $\mu_{1}$ and $\mu_{2}$ be the refractive indices of two media. $v_{1}$ and $v_{2}$ are the velocities of light in the media respectively. Which one of the following relations is $TRUE$?

The apparent depth of water in a cylindrical water tank of diameter $2R \, cm$ is reducing at the rate of $x \, cm/minute$ when water is being drained out at a constant rate. The amount of water drained in $c.c./minute$ is ($n_1 =$ refractive index of air,$n_2 =$ refractive index of water).

The velocity of light in a medium is $1.5 \times 10^8 \ m/s$. Its refractive index will be:

The frequency of a light ray is $6 \times 10^{14} \,Hz$. Its frequency when it propagates in a medium of refractive index $1.5$, will be

The velocity of light in glass,whose refractive index with respect to air is $1.5$,is $2 \times 10^8 \ m/s$. In a certain liquid,the velocity of light is found to be $2.5 \times 10^8 \ m/s$. The refractive index of the liquid with respect to air is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module