Problem Statement
A ray of light in air strikes a glass surface (refractive index $1.50$) at an angle of incidence of $30^\circ$. Find the angle of refraction inside the glass. (HC Verma, Chapter 18 — Geometrical Optics, Problem 77.)
Given Information
- $n_1 = 1.00$ — refractive index of air
- $n_2 = 1.50$ — refractive index of glass
- $\theta_1 = 30^\circ$ — angle of incidence
Physical Concepts & Formulas
At a boundary between two media light bends according to Snell’s law, governed by the ratio of refractive indices. Spherical mirrors and thin lenses image objects through the mirror/lens equation relating object distance, image distance and focal length.
- $n_1\sin\theta_1=n_2\sin\theta_2$ — Snell’s law of refraction
- $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$ — mirror/lens equation
- $m=-\dfrac{v}{u}$ — linear magnification
Step-by-Step Solution
Step 1 — Apply Snell’s law at the interface: The product of refractive index and the sine of the ray angle is conserved across the boundary.
$$n_1\sin\theta_1=n_2\sin\theta_2$$
Step 2 — Solve for the refracted angle: Rearrange to isolate the sine of the transmitted angle.
$$\sin\theta_2=\dfrac{n_1\sin\theta_1}{n_2}$$
Step 3 — Evaluate the angle of refraction: Insert the indices and incidence angle, then take the inverse sine.
$$\sin\theta_2=\dfrac{1.00\times\sin30^\circ}{1.50}$$
Worked Calculation
$$\sin\theta_2=\dfrac{0.5}{1.5}=0.333\quad\Rightarrow\quad \theta_2=19.5^\circ$$
Answer
$$\boxed{\theta_2\approx19.5^\circ}$$
Light entering the denser glass bends toward the normal, slowing and changing direction as it crosses the interface.
Physical Interpretation
Bending from 30 deg to about 19.5 deg is the expected behaviour going from air into ordinary crown glass and underlies how lenses focus light. The denser the second medium, the stronger the bending toward the normal.
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