Question

The figure shows a ray incident at incidence angle ($i$) $\pi /3$. The plot drawn shows the variation of $|r-i|$ versus $({\mu }_1/{\mu }_2=k$), where $r$ = 'angle of refraction'.
The ray is going from ${\mu }_2$ to ${\mu }_1$. Choose possible correct options.

• A. $K_1=\frac{2}{\sqrt{3}}$
• B. ${\theta }_1=\frac{\pi }{6}$
• C. $K_2=1$
• D. ${\theta }_2=\frac{\pi }{3}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B ${\theta }_1=\frac{\pi }{6}$
C $K_2=1$
D ${\theta }_2=\frac{\pi }{3}$

|r - i| = 0 when $K=\frac{{\mu }_1}{{\mu }_2}=1$ So $K_2=1$
When $K\to \infty ,\frac{{\mu }_1}{{\mu }_2}=\infty \Rightarrow r=0So|r-i|=i=\frac{\pi }{3}$
So ${\theta }_2=\frac{\pi }{3}$
Now $K_1<K_2$ so $K_1<1$

If K is smaller than 1, then TIR occurs.
$\Rightarrow {\theta }_1=\left(r-i\right|=\left(90-60\right|=30°=\frac{\pi }{6}$