Question

A ray of light, emitted beneath the surface of an unknown liquid with air above it, undergoes total internal reflection as shown in Figure 25.55. What is the index of refraction for the liquid and its likely identification?

Figure 25.5 A light ray inside a liquid strikes the surface at the critical angle and undergoes total internal reflection.

Short answer

The refractive index of Benzene is shown in the table.

The liquid's refractive index is 1.501, and the substance is benzene.

Step-by-step solution

Definition of angle of refraction

At the point of refraction, the angle formed by a refracted beam and a line drawn normal to the interface between two media.

Calculating the index for the liquid

Let us calculate the index for the liquid.

The expression for critical angle is:

$\sin {\theta }_{\circ }=\frac{n_2}{n_1}..........\left(1\right)$

Where,

• ${\theta }_{\circ }$is critical angle
• $n_2$is the refractive index of air.
• $n_1$is the refractive index of liquid

From the above figure,

$$\begin{gathered} \tan (90-{\theta }_{\circ })=\frac{15}{13.4} \\ cot{\theta }_{\circ }=\frac{15}{13.4} \\ {\theta }_{\circ }=co{t}^{-1}\left(\frac{15}{13.4}\right) \\ {\theta }_{\circ }=41.{78}^{°} \end{gathered}$$

Rearranging the equation,

$n_1=\frac{n_2}{\sin {\theta }_{\circ }}..........\left(2\right)$

Substitute 1 for $n_2=1$and ${\theta }_{\circ }=41.{78}^{°}$in the equation (2),

$$\begin{gathered} n_1=\frac{1}{\sin 41.{78}^{°}} \\ =1.501 \end{gathered}$$

From the table 25.1, the refractive index of Benzene is 1.50.

Therefore, the refractive index of the liquid is 1.501, and the liquid is benzene.