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Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism. The angle of prism is $\left(\cos 41^{\circ}=0.75\right)$
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The correct answer is:
$82^{\circ}$
By prism formula $n=\frac{\sin \frac{A+A}{2}}{\sin \frac{A}{2}}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}$
$\therefore \cos \frac{A}{2}=\frac{n}{2}=\frac{1.5}{2}=0.75=\cos 41^{\circ} \Rightarrow A=82^{\circ}$
$\therefore \cos \frac{A}{2}=\frac{n}{2}=\frac{1.5}{2}=0.75=\cos 41^{\circ} \Rightarrow A=82^{\circ}$
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