A ray of light is travelling through a rectangular glass slab (refractive index $\tfrac{3}{2}$) 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 $\tfrac{4}{3}$) 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.

A ray of light is travelling through a rectangular glass slab (refractive index (3)/(2)) and is incident on th

Question

\tfrac{3}{2}

CBSECBSE Class XII Board 2026Subjective· 3mImportance★★★★★

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Concept understanding — Critical Angle Comparison

Critical Angle Comparison: From Intuition to Precision

Imagine you're standing at the edge of a swimming pool, looking down at a coin at the bottom. The coin appears closer to the surface than it actually is. That's refraction — light bends when it moves from water into air. Now imagine tilting your head so you're looking at the coin from a very shallow angle. At some point, the coin suddenly vanishes. You can't see it anymore, no matter how hard you try. That vanishing point is the critical angle.

The Intuition

Light travels at different speeds in different materials. When it crosses from a denser medium (like water or glass) into a rarer medium (like air), it bends away from the normal (the imaginary line perpendicular to the surface). The larger the angle of incidence (the angle at which light hits the boundary), the more it bends away.

At a certain angle of incidence, the refracted ray bends so much that it runs exactly along the surface — it makes a 90° angle with the normal. That's the critical angle. If you increase the angle of incidence even slightly beyond this, the light can't escape at all. It reflects back into the denser medium, a phenomenon called total internal reflection.

Note

Critical angle only exists when light travels from a denser medium to a rarer medium. Going the other way (rarer to denser), light always bends toward the normal — no critical angle, no total internal reflection.

The Precise Statement

The critical angle ( $\theta_c$ ) is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is exactly 90°.

Mathematically, from Snell's law:

n_1 \sin \theta_1 = n_2 \sin \theta_2

Let medium 1 be the denser medium (refractive index $n_1$ ) and medium 2 be the rarer medium (refractive index $n_2$ , with $n_2 < n_1$ ). At the critical angle, $\theta_1 = \theta_c$ and $\theta_2 = 90^\circ$ , so $\sin \theta_2 = 1$ . This gives:

n_1 \sin \theta_c = n_2 \times 1

\sin \theta_c = \frac{n_2}{n_1}

Where:

$\theta_c$ = critical angle
$n_1$ = refractive index of the denser medium
$n_2$ = refractive index of the rarer medium

What This Tells You

The critical angle depends only on the ratio of the two refractive indices. A larger difference between $n_1$ and $n_2$ means a smaller critical angle — light is more "trapped" inside the denser medium. For example:

Medium pair	$n_1$ (denser)	$n_2$ (rarer)	$\theta_c$
Water → Air	1.33	1.00	$\approx 48.8^\circ$
Glass → Air	1.50	1.00	$\approx 41.8^\circ$
Diamond → Air	2.42	1.00	$\approx 24.4^\circ$

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When the glass slab is in contact with water, the ray refracts at the glass–water interface (no longer at critical angle), then refracts again at the water–air interface. By applying Snell's law twice and using the fact that the original critical angle at glass–air was $\sin^{-1}\!\left(\frac{2}{3}\right)$ , the ray emerges into air at $90°$ to the normal (grazes along the surface).

The heart of this problem is understanding what happens when you insert a third medium between two others. Originally the ray was just at the critical angle for total internal reflection at the glass–air boundary. When water appears between glass and air, the ray now crosses two interfaces: glass → water, then water → air. The critical angle condition at glass–air told us the incident angle; we use that information to track the ray through both refractions.

Why the critical angle matters

At the glass–air interface (before water is introduced), the critical angle $\theta_c$ satisfies

n_{\text{glass}} \sin \theta_c = n_{\text{air}} \sin 90° \quad \Rightarrow \quad \frac{3}{2} \sin \theta_c = 1 \quad \Rightarrow \quad \sin \theta_c = \frac{2}{3}.

This angle $\theta_c$ is the angle of incidence in the glass at which the refracted ray in air would skim along the interface. That's the starting condition: the ray inside the glass is traveling at angle $\theta_c$ to the normal of the horizontal surface.

Step-by-step refraction through glass → water → air

1. Incident angle in glass

The ray hits the glass–water interface at angle $\theta_c$ to the normal, where $\sin \theta_c = \frac{2}{3}$ .

2. Refraction at the glass–water boundary

Snell's law at this interface:

n_{\text{glass}} \sin \theta_c = n_{\text{water}} \sin \theta_1,

where $\theta_1$ is the angle of refraction in water. Substituting the refractive indices:

\frac{3}{2} \cdot \frac{2}{3} = \frac{4}{3} \sin \theta_1 \quad \Rightarrow \quad 1 = \frac{4}{3} \sin \theta_1 \quad \Rightarrow \quad \sin \theta_1 = \frac{3}{4}.

So the ray refracts into the water at an angle $\theta_1 = \sin^{-1}\!\left(\frac{3}{4}\right)$ to the normal.

3. Refraction at the water–air boundary

The ray now travels through the thin water layer and hits the water–air interface at the same angle $\theta_1$ (the interfaces are parallel, so the angle to the normal is preserved). Snell's law at the water–air boundary:

$n_{\text{water}} \sin \theta_1 = n_{\text{air}} \sin \theta_2,$ …

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Why the critical angle matters

Step-by-step refraction through glass → water → air

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