Question

The angle of minimum deviation for prism angle $\frac{\pi }{3}$ is $\frac{\pi }{6}$. Calculate the velocity in the material of the prism, if the velocity of light in vacuum is $3.00\times {10}^{8}m{s}^{-1}$.

• A. $2.12\times {10}^{8}m{s}^{-1}$
• B. $2.50\times {10}^{8}m{s}^{-1}$
• C. $3.00\times {10}^{8}m{s}^{-1}$
• D. $3.50\times {10}^{8}m{s}^{-1}$

Answer 1

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

Draw ray diagram of a light ray passing through the prism



From the triangle, $A+({90}^{\circ }-r_1)++({90}^{\circ }-r_2)={180}^{\circ }$
$\Rightarrow r_1+r_2=A$
And angle of deviation $\delta ={\delta }_1+{\delta }_2$
$=(i-r_1)+(e-r_2)=i+e-A$
Variation of angle of deviation ($\delta$) with angle of incidence ($i$) will look like this



So, $\delta ={\delta }_{min}$ at $i=e$. Also at $i=e$,
$i+e-A=2i-A={\delta }_{min}\Rightarrow i=\frac{A+{\delta }_{min}}{2}$
$r_1+r_2=r+r=2r=A\Rightarrow r=\frac{A}{2}$
Using Snell’s law,
$\mu =\frac{\sin i}{\sin r}=\frac{\sin \left(\frac{A+{\delta }_{min}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Here, $\mu =\frac{{\mu }_P}{{\mu }_M}={\mu }_P$ because the medium is air
So, ${\mu }_P=\frac{\sin \left(\frac{{60}^{\circ }+{30}^{\circ }}{2}\right)}{\sin \left(\frac{{60}^{\circ }}{2}\right)}=\sqrt{2}$
We know ${\mu }_P=\frac{c}{v_P}$
Therefore, velocity of light in prism $v_P=\frac{c}{{\mu }_P}=\frac{3}{\sqrt{2}}\times {10}^8$
$=2.12\times {10}^{8}m/s$