Question

The angle of minimum deviation for prism of angle $\pi /3is\pi /6$, if the velocity of light in vacuum is $3\times {10}^{8}m{s}^{-1}$, then the velocity of light in material of the prism is

• A. $2.12\times {10}^{8}m{s}^{-1}$
• B. $1.12\times {10}^{8}m{s}^{-1}$
• C. $4.12\times {10}^{8}m{s}^{-1}$
• D. $5.12\times {10}^{8}m{s}^{-1}$

Answer 1

The correct option is A $2.12\times {10}^{8}m{s}^{-1}$

Here, $A=\frac{\pi }{3}={60}^{\circ },{\delta }_m=\frac{\pi }{6}={30}^{\circ },c=3\times {10}^{8}m{s}^{-1}$

$\therefore \mu =\frac{sin({60}^{\circ }+{30}^{\circ })/2}{sin\left({60}^{\circ }/2\right)}=\frac{0.7071}{0.50}=1.414$

Therefore, $V=\frac{c}{\mu }=\frac{3\times {10}^8}{1.414}orv=2.12\times {10}^{8}m{s}^{-1}$