Question

Find out no. of reflection after which light ray will exit from (Given sin $$40^0=0.64$$)

Answer 1

Correct option is B. $$57735$$
1 sin $$40^0=1.31sinr$$
$$0.64=1.31sinr$$
$$sinr=\dfrac{0.64}{1.31}=0.49\approx 0.5$$
$$r=30$$
So $$\theta=C$$
$$\because \theta>C$$
$$\therefore$$ T.I.R. at other surface
$$tanr=\dfrac{20\mu m}{x}$$
$$x=20\sqrt{3}$$
$$n=\dfrac{2m}{20\sqrt{3}\mu m}$$
$$n=\dfrac{10^5}{\sqrt{3}}$$
$$n=57735$$