Rankers Physics

Refraction by Prism: Practice Problem & Solution

The angle of incidence for a ray of light at a refracting surface of a prism is $45^\circ$. The angle of prism is $60^\circ$. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are: (2016 - I)
$45^\circ, \frac{1}{\sqrt{2}}$
$30^\circ, \sqrt{2}$
$45^\circ, \sqrt{2}$
$30^\circ, \frac{1}{\sqrt{2}}$

Solution Explained:

To solve this problem, we apply the core principles of Refraction by Prism. Understanding the underlying formula is key to arriving at the correct answer below:

At minimum deviation, angle of emergence equals angle of incidence, so $i = e = 45^\circ$. Minimum deviation $\delta_m = i + e - A = 45^\circ + 45^\circ - 60^\circ = 30^\circ$. Refractive index $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)} = \frac{\sin(45^\circ)}{\sin(30^\circ)} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$.

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