Question

In the figure shown, the light beam strikes the surface $$ 2 $$ at the critical angle. Determine the angle of incidence $$(i_1)$$.

Answer 1

Correct option is A. $$ 27.5^{\circ} $$

If the refractive index of object is $$n_1$$ and that of air outside is $$n_2 = 1$$:

For the refraction at the surface 2:

$$\dfrac{\sin r_2}{\sin i_2} = \dfrac{n_1}{n_2}$$

But for critical angle, $$i_2 = 42^\circ$$, $$r_2 = 90^\circ$$.

$$\Rightarrow n_1 = \dfrac{1}{\sin 42^\circ}$$

Now, angle between the ray inside the object and surface $$2$$ = $$90^\circ - 42^\circ$$ = $$48^\circ$$ (Linear pair)

Then, the angle between the ray inside the object and surface $$1$$ = $$180^\circ - 60^\circ - 48^\circ$$ = $$72^\circ$$ (Since the object forms equilateral triangle)

Thus,

Refraction angle at surface $$1$$, $$r_1 = 90^\circ - 72^\circ = 18^\circ$$

And by Snell's law:

$$\dfrac{\sin r_1}{\sin i_1} = \dfrac{n_2}{n_1}$$

$$\Rightarrow \sin i_1 = n_1 \sin r_1 $$

$$\Rightarrow \sin i_1 = \dfrac{\sin 18^\circ}{\sin 42^\circ}$$

$$\Rightarrow i_1 = \sin^{-1} \left(\dfrac{\sin 18^\circ}{\sin 42^\circ}\right)$$

$$\therefore i_1 = 27.5^\circ$$