Question

An isosceles prism of angle 120° has a refractive index 1.44. Two parallelmonochromatic rays enter the prism parallel to each other in air as shown. The rays emerge from the opposite face-
(A) are parallel to each other
(B) are diverging
(C) make an angle 2 [sin−1(0.72) − 30°] with each other
(D) make an angle 2 sin −1(0.72) with each other

Answer 1

Answer on Question #49689-Physics-Optics

An isosceles prism of angle $120{}^{\circ}$ has a refractive index 1.44. Two parallel monochromatic rays enter the prism parallel to each other in air as shown. The rays emerge from the opposite face-

(A) are parallel to each other

(B) are diverging

(C) make an angle 2 [sin-1(0.72) - 30°] with each other

(D) make an angle 2 sin -1(0.72) with each other

Solution

As the light rays are incident normally on the first interface so they do not refract and strike the inclined faces of prism at an angle $30{}^{\circ}$ .

Thus, Incident angle $i = 30{}^{\circ}$ .

So by using Snell's law we get,

$$\frac {\sin i}{\sin r} = \frac {n _ {2}}{n _ {1}} \rightarrow \frac {\sin 3 0}{\sin r} = \frac {1}{1 . 4 4} \rightarrow r = \sin^ {- 1} 0. 7 2.$$

Now in quadrilateral ABCD, from above figure we get,

$$\angle A + \angle B + \angle C + \angle D = 3 6 0 {}^ {\circ} \rightarrow (1 8 0 {}^ {\circ} - r) + x + (1 8 0 {}^ {\circ} - r) + 6 0 {}^ {\circ} = 3 6 0 {}^ {\circ}.$$

By solving, we get

$$x = 2 r - 6 0 {}^ {\circ} = 2 \left(\sin^ {- 1} 0. 7 2 - 3 0 {}^ {\circ}\right).$$

Answer: (C) make an angle $2\left(\sin^{-1}0.72 - 30{}^{\circ}\right)$ with each other.

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