Given small samples of three liquids, you are asked to determine their refractive indexes. However, you do not have enough of each liquid to measure the angle of refraction for light refracting from air into the liquid. Instead, for each liquid, you take a rectangular block of glass 1n= 1.522 and place a drop of the liquid on the top surface of the block. You shine a laser beam with wavelength 638 nm in vacuum at one side of the block and measure the largest angle of incidence uafor which there is total internal reflection at the interface between the glass and the liquid (Fig. P33.58). Your results are given in the table: Liquid A B C

Ua1 _2 52.0 44.3 36.3 What is the refractive index of each liquid at this wavelength?

First, we apply Snell's law for each liquid in the form

$n_a\sin {\theta }_a=n_b\sin {\theta }_b$, (1)

Where $n_a$is the refractive index of the medium with the incident light, ${\theta }_a$is the incident angle, $n_b$, is the refractive index of the medium with the refractive beam, and ${\theta }_b$, is the refractive angle. At the interface between liquid and the glass, the angle of the refraction is $\theta$. So, it is related to ${\theta }_b$, by

$\theta =90°-{\theta }_b$(2)

For liquid A with a refractive index $n_a$apply Snell's law for the interface between the air and the glass to get ${\theta }_b$, as next

$$\begin{gathered} n_a\sin {\theta }_a=n_b\sin {\theta }_b \\ \left(1\right)\sin \left(52.0°\right)=\left(1.52\right)\sin {\theta }_b \\ \left(1\right)\left(0.788\right)=\left(1.52\right)\sin {\theta }_b \\ \sin {\theta }_b=\left(0.518\right) \\ {\theta }_b={\sin }^{-1}\left(0.518\right) \\ {\theta }_b=31.23° \end{gathered}$$

Use equation (2) to get$\theta$ as next

$\theta =90°-31.23°=58.77°$

Now, apply Snell's law for the interface between the liquid and the glass to get$n_a$ as next

$$\begin{gathered} n\sin \theta =n_A\sin 90° \\ \left(1.52\right)\sin 58.77°=n_A \\ {n}_A=1.30 \end{gathered}$$

For liquid B with refractive index $n_a$apply Snell's law for the interface between the air and the glass to get${\theta }_b$ , as next

$$\begin{gathered} n_a\sin {\theta }_a=n_b\sin {\theta }_b \\ \left(1\right)\sin \left(43.3°\right)=\left(1.52\right)\sin {\theta }_b \\ \sin {\theta }_b=\left(0.459\right) \\ {\theta }_b={\sin }^{-1}\left(0.459\right) \\ {\theta }_b=27.35° \end{gathered}$$

Use equation (2) to get$\theta$ as next

$\theta =90°-27.35°=62.65°$

Now, apply Snell's law for the interface between the liquid and the glass to get$n_b$as next

$$\begin{gathered} n\sin \theta =n_B\sin 90° \\ \left(1.52\right)\sin 62.65°=n_B \\ {n}_B=1.35 \end{gathered}$$

For liquid C with a refractive index $n_b$apply Snell's law for the interface between the air and the glass to get${\theta }_b$ , as next

$$\begin{gathered} n_a\sin {\theta }_a=n_b\sin {\theta }_b \\ \left(1\right)\sin \left(36.3°\right)=\left(1.52\right)\sin {\theta }_b \\ \sin {\theta }_b=\left(0.389\right) \\ {\theta }_b={\sin }^{-1}\left(0.389\right) \\ {\theta }_b=22.92° \end{gathered}$$

Use equation (2) to get$\theta$ as next

$\theta =90°-22.92°=67.08°$

Now, apply Snell's law for the interface between the liquid and the glass to getas next

$$\begin{gathered} n\sin \theta =n_c\sin 90° \\ \left(1.52\right)\sin 67.08°=n_B \\ {n}_c=1.40 \end{gathered}$$