Refraction by Prism: Practice Problem & Solution
For a prism its refractive index is $\cot A/2$ then minimum angle of deviation is: (1999)
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:
Refractive index $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$. Given $\mu = \cot(A/2) = \frac{\cos(A/2)}{\sin(A/2)} = \frac{\sin(90^\circ - A/2)}{\sin(A/2)}$. Equating numerators: $\frac{A + \delta_m}{2} = 90^\circ - \frac{A}{2}$. This gives $A + \delta_m = 180^\circ - A$, so the minimum angle of deviation is $\delta_m = 180^\circ - 2A$.
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