Question

Question: The index of refraction of benzene is$\mathbf{1}\mathbf{.}\mathbf{8}$. What is the critical angle for a light ray traveling in benzene toward a flat layer of air above the benzene?

Short answer

The critical angle for the light ray is ${\theta }_b=33.7^{\circ }\approx {34}^{\circ }$.

Step-by-step solution

Given

The refractive index is $n_b=1.8$.

Determine the concept for the Snell’s law as:

Write for the angle of refraction in terms of the vertex angle from the geometry of the prism and the ray diagram. Then using the formula for the exterior angle of triangle, solve for the angle of incidence in terms of the vertex angle. Then, using Snell’s law of refraction, show that the index of refraction of the glass prism is:

$\mathbf{n}\mathbf{=}\frac{\mathbf{sin}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\mathbf{\psi }\mathbf{+}\mathbf{f}\mathbf{)}}{\mathbf{sin}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{f}}$

Formula:

${\mathrm{n}}_1{\mathrm{sin\theta }}_1={\mathrm{n}}_2{\mathrm{sin\theta }}_2$

Calculate the critical angle for a light ray traveling in benzene toward a flat layer of air above the benzene

Snell’s law of refraction gives that

$$\begin{gathered} {\mathrm{n}}_1{\mathrm{sin\theta }}_1={\mathrm{n}}_2{\mathrm{sin\theta }}_2 \\ {\mathrm{n}}_{\mathrm{b}}{\mathrm{sin\theta }}_{\mathrm{a}}={\mathrm{n}}_{\mathrm{a}}{\mathrm{sin\theta }}_{\mathrm{a}} \\ {\mathrm{sin\theta }}_{\mathrm{b}}=\frac{{\mathrm{n}}_{\mathrm{a}}}{{\mathrm{n}}_{\mathrm{b}}}{\mathrm{sin\theta }}_{\mathrm{a}} \\ {\mathrm{sin\theta }}_{\mathrm{b}}=\frac{1}{1.8}\sin {90}^{\circ } \end{gathered}$$

Substitute the values and solve as:

$\begin{array}{ll}{\mathrm{sin\theta }}_{\mathrm{b}}=0.55& \\ {\mathrm{\theta }}_{\mathrm{b}}=33.7^{\circ }\approx {34}^{\circ }& \end{array}$