Angle of minimum deviation for a prism of ref | vedclass

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(C) The formula for the refractive index of a prism is $\mu = \frac{\sin \left(\frac{\delta_{\min} + A}{2}\right)}{\sin(A/2)}$.Given that the angle of

Vedclass · Accepted Answer

$82$

Vedclass · Answer

(C) The formula for the refractive index of a prism is $\mu = \frac{\sin \left(\frac{\delta_{\min} + A}{2}\right)}{\sin(A/2)}$.Given that the angle of minimum deviation $\delta_{\min} = A$,we substitute this into the formula:$\mu = \frac{\sin \left(\frac{A + A}{2}\right)}{\sin(A/2)} = \frac{\sin A}{\sin(A/2)}$.Using the trigonometric identity $\sin A = 2 \sin(A/2) \cos(A/2)$,we get:$\mu = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)} = 2 \cos(A/2)$.Given $\mu = 1.5 = \frac{3}{2}$,we have:$\frac{3}{2} = 2 \cos(A/2) \Rightarrow \cos(A/2) = \frac{3}{4} = 0.75$.Since $\cos 41^{\circ} = 0.75$,it follows that $A/2 = 41^{\circ}$.Therefore,$A = 82^{\circ}$.

Angle of minimum deviation for a prism of refractive index $1.5$ is equal to the angle of prism. The angle of prism is $......^{\circ}$ $\left(\cos 41^{\circ}=0.75\right)$

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