91Ó°ÊÓ

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

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

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

$\mathrm{θ}=90°-{\mathrm{θ}}_b$(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 ${\mathrm{θ}}_b$, as next

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

Use equation (2) to get$\mathrm{θ}$ as next

$\mathrm{θ}=90°-31.23°=58.77°$

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

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

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

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

Use equation (2) to get$\mathrm{θ}$ as next

$\mathrm{θ}=90°-27.35°=62.65°$

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

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

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

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

Use equation (2) to get$\mathrm{θ}$ as next

$\mathrm{θ}=90°-22.92°=67.08°$

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

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