Question

If minimum deviation = $30^{\circ }$ then speed of light in prism is

• A. $\frac{3}{\sqrt{2}}\times 10^8$ m/s
• B. $\frac{2}{\sqrt{3}}\times 10^8$ m/s
• C. $\frac{1}{\sqrt{2}}\times 10^8$ m/s
• D. $\frac{2}{3}\times 10$ m/s

Answer 1

Answer: (A)

We know the prism formula,
$\mu =\frac{\sin \left(\frac{(A+\delta m)}{2}\right)}{\sin (A/2)}$
But $A=60^{\circ }8m=30^{\circ }\to$ [Given]
$\therefore \mu =\frac{\sin \left(\frac{60+30}{2}\right)}{\sin (60/2)}=\frac{\sin 45}{\sin 30}$
$\therefore \mu =\sqrt{2}$
But $\mu =\frac{\text{ speed of light in air (c) }}{\text{ speed of ligut in prism (v) }}$
$\therefore v=\frac{c}{u}=\frac{3\times 10^8}{\sqrt{2}}$
$=\frac{3}{\sqrt{2}}\times 10^{8}m/s$