Question

If speed of light is \(2.12\times10^8\,\text{m/s}\) in the medium of an equilateral prism, find the minimum angle of deviation.

• A. \(45^\circ\)
• B. \(60^\circ\)
• C. \(30^\circ\)
• D. \(53^\circ\)

Answer 1

The Correct Option is C

Concept: Refractive index: \[ \mu=\frac{c}{v} \] Minimum deviation relation: \[ \mu=\frac{\sin\left(\frac{\delta_{min}+A}{2}\right)}{\sin\left(\frac{A}{2}\right)} \] Step 1: Find refractive index \[ \mu=\frac{3\times10^8}{2.12\times10^8} \] \[ \mu=\sqrt{2} \] Step 2: Use prism formula For equilateral prism \[ A=60^\circ \] \[ \sqrt{2}=\frac{\sin\left(\frac{\delta_{min}+60^\circ}{2}\right)}{\sin30^\circ} \] \[ \sin30^\circ=\frac12 \] \[ \sqrt{2}=2\sin\left(\frac{\delta_{min}+60^\circ}{2}\right) \] \[ \sin\left(\frac{\delta_{min}+60^\circ}{2}\right)=\frac{1}{\sqrt{2}} \] Step 3: Solve \[ \frac{\delta_{min}+60^\circ}{2}=45^\circ \] \[ \delta_{min}+60^\circ=90^\circ \] \[ \delta_{min}=30^\circ \] Thus \[ \boxed{30^\circ} \]