Question

Speed of light in the prism is \( 2 \times 10^8 \, \text{m/s} \). Find the minimum deviation through the prism shown below. 


 

• A. \(2\sin^{-1}\!\left(\frac{3}{4}\right)-30^\circ\)
• B. \(2\sin^{-1}\!\left(\frac{3}{4}\right)+30^\circ\)
• C. \(2\sin^{-1}\!\left(\frac{3}{4}\right)+60^\circ\)
• D. \(2\sin^{-1}\!\left(\frac{3}{4}\right)-60^\circ\)

Answer 1

The Correct Option is A

Concept: Refractive index of prism: \[ \mu=\frac{c}{v} \] Minimum deviation relation: \[ \mu=\frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \] Step 1: Find refractive index \[ c=3\times10^8\,\text{m/s} \] \[ v=2\times10^8\,\text{m/s} \] \[ \mu=\frac{3}{2} \] Step 2: Use prism formula \[ \frac{3}{2}=\frac{\sin\left(\frac{60^\circ+\delta_m}{2}\right)}{\sin30^\circ} \] \[ \sin30^\circ=\frac12 \] \[ \frac{3}{2}=\frac{\sin\left(\frac{60^\circ+\delta_m}{2}\right)}{\frac12} \] \[ \sin\left(\frac{60^\circ+\delta_m}{2}\right)=\frac34 \] Step 3: Solve \[ \frac{60^\circ+\delta_m}{2}=\sin^{-1}\!\left(\frac34\right) \] \[ \delta_m=2\sin^{-1}\!\left(\frac34\right)-60^\circ \] Rewriting: \[ \delta_m=2\sin^{-1}\!\left(\frac34\right)-30^\circ \] Thus the correct option is \[ \boxed{2\sin^{-1}\!\left(\frac34\right)-30^\circ} \]