Question

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

• A. $\frac{3}{\sqrt{2}}\times10^{8}$ m/s
• B. $\frac{2}{\sqrt{3}}\times10^{8}$ m/s
• C. $\frac{1}{\sqrt{2}}\times10^{8}$ m/s
• D. $\frac{2}{3}\times10$ 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 $A=60^{\circ} \,\,\, 8 m=30^{\circ} \rightarrow$ [Given]
$\therefore \mu=\frac{\sin \left(\frac{60+30}{2}\right)}{\sin (60 / 2)}=\frac{\sin 45}{\sin 30}$
$\therefore \mu=\sqrt{2}$
But $\mu=\frac{\text { speed of light in air (c) }}{\text { speed of ligut in prism (v) }}$
$\therefore v=\frac{c}{u}=\frac{3 \times 10^{8}}{\sqrt{2}}$
$=\frac{3}{\sqrt{2}} \times 10^{8} m / s$