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A ray of light falling at an angle of 50

Question :- A ray of light falling at an angle of 50° is refracted through a prism and suffers minimum deviation. The angle of the prism is 60° . Find the angle of minimum deviation and refractive index of the material of the prism.

Solution :-

Here i₁ = 50°, A = 60°, δ_m = ? and μ = ?

In case of minimum deviation, i₁ = i₂ = i = 50°

Using δ_m= 2i – A

⇒ δ_m= 2(50°) – 60° = 40°

Now using $\displaystyle \mu =\frac{\sin \left( \frac{A+{{\delta }_{\min }}}{2} \right)}{\sin \left( \frac{A}{2} \right)}$ , we get

$\displaystyle \mu =\frac{\sin \left( \frac{A+{{\delta }_{\min }}}{2} \right)}{\sin \left( \frac{A}{2} \right)}=\frac{\sin \left( \frac{{{60}^{\circ }}+{{40}^{\circ }}}{2} \right)}{\sin \left( \frac{{{60}^{\circ }}}{2} \right)}=\frac{\sin {{50}^{\circ }}}{\sin {{30}^{\circ }}}=\frac{0.7660}{0.5000}$

$\displaystyle \Rightarrow \mu =1.532$

A ray of light falling at an angle of 50

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