The refractive index of a prism is mu = sqrt{ | vedclass

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(A) For minimum deviation $\delta_{\min}$,we have $i = e$ and $r_1 = r_2 = \frac{A}{2}$.Given that the ratio of the angle of minimum deviation $\delta

Vedclass · Accepted Answer

$60$

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(A) For minimum deviation $\delta_{\min}$,we have $i = e$ and $r_1 = r_2 = \frac{A}{2}$.Given that the ratio of the angle of minimum deviation $\delta_{\min}$ to the angle of the prism $A$ is $1$,so $\frac{\delta_{\min}}{A} = 1$,which implies $\delta_{\min} = A$.We know that $\delta_{\min} = 2i - A$. Substituting $\delta_{\min} = A$,we get $A = 2i - A$,which simplifies to $2A = 2i$,or $i = A$.Using Snell's law at the first surface: $1 	imes \sin i = \mu \sin r_1$.Substituting $i = A$ and $r_1 = \frac{A}{2}$,we get $\sin A = \mu \sin \left(\frac{A}{2}ight)$.Using the trigonometric identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$,we have $2 \sin \frac{A}{2} \cos \frac{A}{2} = \sqrt{3} \sin \left(\frac{A}{2}ight)$.Dividing both sides by $\sin \left(\frac{A}{2}ight)$ (since $A 
eq 0$),we get $2 \cos \left(\frac{A}{2}ight) = \sqrt{3}$,or $\cos \left(\frac{A}{2}ight) = \frac{\sqrt{3}}{2}$.This implies $\frac{A}{2} = 30^{\circ}$,so $A = 60^{\circ}$.

The refractive index of a prism is $\mu = \sqrt{3}$ and the ratio of the angle of minimum deviation to the angle of the prism is $1$. The value of the angle of the prism is $......^{\circ}$.

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