Question

A ray of light is travelling through a rectangular glass slab (refractive index \(\frac{3}{2}\)) and is incident on the horizontal glass-air surface at the critical angle for the two media. The slab is then brought in contact with water (refractive index \(\frac{4}{3}\)) such that a thin horizontal layer of water is formed on the surface of the slab. Find the angle at which the ray will emerge into air from the water-air surface.

Hint:

For parallel layer media arrangements, intermediate layers do not alter the final output angle relations between the first and last boundaries. Since the light entered at the critical angle for glass-air, it will exit at \(90^{\circ}\) into the air, regardless of what intermediate layers like water are added in between.

Answer 1

Concept: According to Snell's Law of refraction, when a ray of light passes across multiple parallel plane interfaces separating different media, the product of the refractive index and the sine of the angle of incidence/refraction remains constant throughout the path: \[ \mu_1 \sin\theta_1 = \mu_2 \sin\theta_2 = \mu_3 \sin\theta_3 = \text{Constant} \] The critical angle \(\theta_c\) for a boundary separating a denser medium of index \(\mu_d\) and a rarer medium of index \(\mu_r\) is defined by: \[ \sin\theta_c = \frac{\mu_r}{\mu_d} \]

Step 1: Analyzing the initial glass-air configuration.
Initially, the ray travels in glass (\(\mu_g = \frac{3}{2}\)) and is incident on the glass-air boundary at the critical angle \(\theta_c\). The surrounding medium is air (\(\mu_a = 1\)). \[ \sin\theta_c = \frac{\mu_a}{\mu_g} = \frac{1}{3/2} = \frac{2}{3} \]

Step 2: Analyzing the system after adding the water layer.
Now, a horizontal layer of water (\(\mu_w = \frac{4}{3}\)) is placed on top of the glass slab. The ray originates with the exact same initial angle of incidence \(\theta_c\) within the glass. It refracts first at the glass-water interface at an angle \(\theta_w\), and then arrives at the water-air interface, finally emerging into the air at an angle \(e\). Applying continuous parallel layer forms of Snell's Law across all three successive media (Glass \(\rightarrow\) Water \(\rightarrow\) Air): \[ \mu_g \sin\theta_c = \mu_w \sin\theta_w = \mu_a \sin e \]

Step 3: Calculating the angle of emergence \(e\).
By equating the expression for the first medium (glass) directly with the final medium (air): \[ \mu_g \sin\theta_c = \mu_a \sin e \] Substituting our known values (\(\mu_g = \frac{3}{2}\), \(\sin\theta_c = \frac{2}{3}\), and \(\mu_a = 1\)): \[ \left(\frac{3}{2}\right) \times \left(\frac{2}{3}\right) = 1 \times \sin e \] \[ 1 = \sin e \] Since \(\sin e = 1\), the angle of emergence must be: \[ e = 90^{\circ} \] This means the ray of light will graze along the final horizontal water-air boundary line as it exits.