Rankers Physics

Refraction by Prism: Practice Problem & Solution

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index: (2012 Mains)
Lies between $\sqrt{2}$ and $1$
Lies between $2$ and $\sqrt{2}$
Is less than $1$
Is greater than $2$

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:

Given $\delta_m = A$. The refractive index is $\mu = \frac{\sin((A+A)/2)}{\sin(A/2)} = \frac{\sin A}{\sin(A/2)} = 2\cos(A/2)$. For a physical prism, the maximum angle of incidence is $90^\circ$, which corresponds to $A = 90^\circ$ for $\delta_m = A$, giving $\mu = 2\cos(45^\circ) = \sqrt{2}$. As $A \to 0$, $\mu \to 2$, so the refractive index lies between $2$ and $\sqrt{2}$.

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