The exit surface of a prism with refractive i | vedclass

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Question

(A) For the condition of minimum deviation,the angle of incidence $i$ is equal to the angle of emergence $e$,and the angle of refraction $r$ is given

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) For the condition of minimum deviation,the angle of incidence $i$ is equal to the angle of emergence $e$,and the angle of refraction $r$ is given by $r = \frac{A}{2}$,where $A$ is the prism angle.At the exit surface,the light ray strikes the interface between the prism (refractive index $n$) and the coating (refractive index $n/2$).For the condition of the critical angle $	heta_{c}$,the angle of refraction $r$ must be equal to $	heta_{c}$.According to Snell's law at the exit surface: $n \sin(r) = (n/2) \sin(90^{\circ})$.Since $\sin(90^{\circ}) = 1$,we have $n \sin(r) = n/2$.Dividing both sides by $n$,we get $\sin(r) = 1/2$.Since $r = A/2$,we have $\sin(A/2) = 1/2$.Therefore,$A/2 = 30^{\circ}$,which gives $A = 60^{\circ}$.

The exit surface of a prism with refractive index $n$ is coated with a material having refractive index $\frac{n}{2}$. When this prism is set for minimum angle of deviation,it exactly meets the condition of the critical angle. The prism angle is . . . . . . (in $^{\circ}$)

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