A ray of light passing through a prism (mu = | vedclass

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(D) At minimum deviation,the angle of refraction $r_1 = r_2 = r = \frac{A}{2}$.Given that the angle of incidence $i$ is double the angle of refraction

Vedclass · Accepted Answer

$60$

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(D) At minimum deviation,the angle of refraction $r_1 = r_2 = r = \frac{A}{2}$.Given that the angle of incidence $i$ is double the angle of refraction $r$,so $i = 2r = A$.According to Snell's Law,$1 \cdot \sin i = \mu \cdot \sin r$.Substituting the values,$\sin A = \sqrt{3} \sin \frac{A}{2}$.Using the trigonometric identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$,we get $2 \sin \frac{A}{2} \cos \frac{A}{2} = \sqrt{3} \sin \frac{A}{2}$.Since $\sin \frac{A}{2} 
eq 0$,we have $\cos \frac{A}{2} = \frac{\sqrt{3}}{2}$.This implies $\frac{A}{2} = 30^{\circ}$,so $A = 60^{\circ}$.

$A$ ray of light passing through a prism $(\mu = \sqrt{3})$ suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then,the angle of prism is ..... (in degrees).

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