Rankers Physics

Refraction by Prism: Practice Problem & Solution

For a prism its refractive index is $\cot A/2$ then minimum angle of deviation is: (1999)
$180 - A$
$180 - 2A$
$90 - A$
$A/2$

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:

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 $\delta_m = 180^\circ - 2A$.

Leave a Reply

Your email address will not be published. Required fields are marked *