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Q. If refractive index of a material of equilateral prism is $ \sqrt{3}, $ then angle of minimum deviation of the prism is

ManipalManipal 2009Ray Optics and Optical Instruments

Solution:

$\mu =\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \frac{A}{2}}$
$\sqrt{3}=\frac{\sin \left(\frac{60^{\circ}+\delta_{m}}{2}\right)}{\sin \frac{60^{\circ}}{2}}$
$\Rightarrow \frac{\sqrt{3}}{2}=\sin \left[30^{\circ}+\frac{\delta_{m}}{2}\right]$
$\Rightarrow \delta_{m}=60^{\circ}$