Question

A layer of oil of refractive index $$ n_o $$ floats on water of refractive index $$ n_w $$. A beam of light is directed upward from a source in the water. Show that the critical angle for a total internal reflection is independent of $$ n_o $$, even though, for angles close to C (critical angle), the total internal reflection always takes place at the oil-air surface.

Answer 1

Let the incidence angle at water-oil interface be $$i$$, and the refraction angle be $$r$$, which also acts as the incidence angle at oil-air interface.

Then, at the water-oil interface by Snell's law:

$$\dfrac{\sin r}{\sin i} = \dfrac{n_w}{n_o}$$

$$\Rightarrow \sin r = \dfrac{n_w}{n_o}\sin i$$

Again, for Total internal reflection at oil-air interface, by Snell's law:

$$\dfrac{\sin 90^{\circ}}{\sin r}=\dfrac{n_o}{1}$$ (since, $$n_{air} \approx 1 $$)

$$\Rightarrow \sin r = \dfrac{1}{n_o}$$

$$\Rightarrow \dfrac{n_w}{n_o}\sin i = \dfrac{1}{n_o}$$

$$\Rightarrow \sin i = \dfrac{1}{n_w}$$

$$\therefore i = \sin^{-1}\left(\dfrac{1}{n_w}\right)=C$$

Thus, the critical angle $$C$$ for total internal reflection is independent of $$n_o$$.