First slide

Prism

Question

When light of wavelength $λ$ is incident on an equilateral prism kept in its minimum deviation position, it is found that the angle of deviation equals the angle of the prism itself. The refractive index of the material of the prism for the wavelength $λ$ is, then

Moderate

A

$\sqrt{3}$
B

$\frac{\sqrt{3}}{2}$
C

2
D

$\sqrt{2}$

Solution

$μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}} = \frac{\sin (\frac{60^{\circ} + 60^{\circ}}{2})}{\sin (\frac{60^{\circ}}{2})} = \sqrt{3}$

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