Question

A prism is made up of material of refractive index $\sqrt{2}$. The angle of the prism is $A$. If the angle of minimum deviation is equal to the angle of the prism, the value of $A$ is

• A. $30^\circ$
• B. $45^\circ$
• C. $60^\circ$
• D. $75^\circ$
• E. $90^\circ$

Hint:

For minimum deviation problems where $\delta_m = A$, the formula always simplifies to $n = 2 \cos(A/2)$. This is a very common special case in competitive exams.

Answer 1

Answer: (E)

Concept:
The prism formula relating refractive index ($n$), prism angle ($A$), and minimum deviation ($\delta_m$) is: \[ n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

Step 1: Substitute the given conditions.
Given $n = \sqrt{2}$ and $\delta_m = A$: \[ \sqrt{2} = \frac{\sin\left(\frac{A + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin A}{\sin\left(\frac{A}{2}\right)} \]

Step 2: Solve for $A$ using the double-angle identity.
Recall $\sin A = 2 \sin(A/2) \cos(A/2)$: \[ \sqrt{2} = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)} \] \[ \sqrt{2} = 2 \cos(A/2) \implies \cos(A/2) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \] \[ A/2 = 45^\circ \implies A = 90^\circ \]