First slide

Ray optics

Question

The angle of a prism is 60° and its refractive index is $\sqrt{2}$ The angle of minimum deviation suffered by a ray of light in passing through it is

Easy

A

About 20°
B

30°
C

60°
D

45°

Solution

$\begin{array}{l} μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})} \Rightarrow \sqrt{2}; μ = \frac{\sin (\frac{60 + δ_{m}}{2})}{\sin (\frac{60}{2})} \\ \Rightarrow \sqrt{2} \times \sin 30 = \sin (\frac{60 + δ_{m}}{2}) \Rightarrow \sin 45^{o} \\ = \sin (\frac{60 + δ_{m}}{2}) \Rightarrow d_{m} = 30^{o} \end{array}$

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