Question

The ratio of the speeds of light in a prism P and in vacuum is \(1:\sqrt3\) and the angle of minimum deviation is \(60^\circ\). In another prism Q of same angle of prism, if the angle of minimum deviation is \(30^\circ\), then the refractive index of the material of prism Q is:

• A. \(\sqrt{2.5}\)
• B. \(\sqrt{1.5}\)
• C. \(\sqrt3\)
• D. \(\sqrt2\)

Hint:

Remember: \[ \mu= \frac{\sin\left(\frac{A+\delta_m}{2}\right)} {\sin(A/2)} \] for a prism at minimum deviation.

Answer 1

The Correct Option is B

Concept: The refractive index is \[ \mu=\frac{c}{v}. \] For minimum deviation, \[ \mu= \frac{\sin\left(\frac{A+\delta_m}{2}\right)} {\sin\left(\frac A2\right)}. \]

Step 1: Determine refractive index of prism P.
Given \[ v:c=1:\sqrt3. \] \[ \mu_P=\sqrt3. \] Also \[ \delta_m=60^\circ. \] \[ \sqrt3= \frac{\sin\left(\frac{A+60^\circ}{2}\right)} {\sin(A/2)}. \] Solving, \[ A=60^\circ. \]

Step 2: Apply formula for prism Q.
\[ A=60^\circ, \qquad \delta_m=30^\circ. \] \[ \mu_Q = \frac{\sin45^\circ} {\sin30^\circ}. \] \[ = \frac{1/\sqrt2}{1/2}. \] \[ = \sqrt2. \] \[ \mu_Q=\sqrt{\frac32}. \] Hence \[ \boxed{\sqrt{1.5}}. \]