Question

Find the refractive index of material of a prism, if the angle of prism and the angle of minimum deviation are equal to $A$.

Hint:

For $D_m = A$, $\mu = 2\cos\left(\tfrac{A}{2}\right)$ is a useful shortcut.

Answer 1

Step 1: Formula for refractive index of prism.
\[ \mu = \frac{\sin \left(\frac{A + D_m}{2}\right)}{\sin \left(\frac{A}{2}\right)} \] where $A$ = angle of prism, $D_m$ = angle of minimum deviation.
Step 2: Substitution.
Given $D_m = A$, \[ \mu = \frac{\sin \left(\frac{A + A}{2}\right)}{\sin \left(\frac{A}{2}\right)} = \frac{\sin A}{\sin \left(\frac{A}{2}\right)} \]
Step 3: Trigonometric simplification.
Using $\sin A = 2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)$, \[ \mu = \frac{2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)} = 2\cos \left(\frac{A}{2}\right) \]
Step 4: Condition for minimum deviation $A = D_m$.
This condition holds when $A = 60^\circ$, so \[ \mu = 2\cos 30^\circ = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}. \] If $A = 90^\circ$, then \[ \mu = 2\cos 45^\circ = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}. \]
Step 5: Conclusion.
Depending on $A$, the refractive index comes out to be $\sqrt{2}$ (commonly used case).