First slide

Prism

Question

The light ray is incidence at angle of 60° on a prism of angle 45°. When the light ray falls on the other surface at 90°, the refractive index of the material of prism $μ$ and the angle of deviation $δ$ are given by

Moderate

A

$μ = \sqrt{2}, δ = 30^{\circ}$
B

$μ = 1 .5, δ = 15^{\circ}$
C

$μ = \frac{\sqrt{3}}{2}, δ = 30^{\circ}$
D

$μ = \sqrt{\frac{3}{2}}, δ = 15^{\circ}$

Solution

From figure it is clear that $∠ e = ∠ r_{2} = 0$
$From A = r_{1} + r_{2}$
$\Rightarrow r_{1} = A = 45^{\circ}$
$∴ μ = \frac{sini}{{sinr}_{1}} = \frac{\sin 60}{\sin 45} = \sqrt{\frac{3}{2}}$
Also
from $i + e = A + δ \Rightarrow 60 + 0 = 45 + δ \Rightarrow δ = 15^{\circ}$

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