Question

The figure shows a ray incident at an angle angle $i=\pi /3.$ if the plot drawn shown the variation i ( versus $\frac{{\mu }_1}{{\mu }_2}=k,$(r=angle of refraction )

• A. the value of $k_{1}is\frac{2}{\sqrt{3}}$
• B. the value of ${\theta }_2=\pi /3$
• C. the value of ${\theta }_2=\pi /6$
• D. the value of $K_{2}is1$

Answer 1

Answer: (B), (D)

The correct options are
B the value of ${\theta }_2=\pi /3$
D the value of $K_{2}is1$

Given,

The angle of incidence $i={60}^{\circ }$

The ratio of refractive indices $\frac{{\mu }_1}{{\mu }_2}=k$

From the given figure, using Snell’s law

${\mu }_2\sin i={\mu }_1\sin r$

$\frac{{\mu }_1}{{\mu }_2\sin {60}^{\circ }}=\frac{1}{\sin r}\Rightarrow \sin r=\frac{\sqrt{3}}{2k}$ ……. (1)

Substitute the value of $k=1$ in equation (1)

$\sin r=\frac{\sqrt{3}}{2}\Rightarrow r={\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Hence, the angle of refraction is$r=\frac{\pi }{3},\frac{2\pi }{3}$

Now substituting the value of $r=\frac{2\pi }{3}$ in the given equation ${\theta }_2=|r-i|$

${\theta }_2=|r-i|=\left|\frac{2\pi }{3}-\frac{\pi }{3}\right|=\frac{\pi }{3}$

Hence the angle ${\theta }_2=\frac{\pi }{3}$ and $k_2=1$

Therefore, the option (B) and (D) is the correct answer.