A ray of light $P Q$ is incident on the face $A B$ of an equilateral glass prism $A B C$ as shown in the figure. Show that angle of emergence=angle of deviation. Also, with a ray diagram, trace the path of the emergent ray.

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Solution

Tracing the path of an emergent ray of the given diagram,
Now, $i = 60^{o}$ and $∠ B = 60^{o}$
We know, $i + e = B + D, w h e r e i = a n g l e o f i n c i d e n c e, e = a n g l e o f e m e r g e n c e, B = a n g l e o f p r i s m, D = a n g l e o f d e v i a t i o n$
As,
$i = B \Rightarrow e = D$
Thus, It is proved that the angle of emergence is equal to the angle of deviation.

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A ray of light PQ is incident on the face AB of an equilateral glass prism ABC as shown in the figure. Show that angle of emergence=angle of deviation. Also, with a ray diagram, trace the path of the emergent ray.

Tracing the path of an emergent ray of the given diagram, Now, i=60o and ∠B=60oWe know, i+e=B+D,wherei=angleofincidence,e=angleofemergence,B=angleofprism,D =angleofdeviationAs,i=B⇒e=DThus, It is proved that the angle of emergence is equal to the angle of deviation.

A ray of light $P Q$ is incident on the face $A B$ of an equilateral glass prism $A B C$ as shown in the figure. Show that angle of emergence=angle of deviation. Also, with a ray diagram, trace the path of the emergent ray.