The refractive indices of two media A and B are $2$ and $\sqrt{2}$ respectively. What is the critical angle for their interface ?

The refractive indices of two media A and B are 2 and √(2) respectively. What is the critical angle for their

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CBSECBSE Class XII Board 2023Subjective· 2mImportance★★★★★

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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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The critical angle occurs when light travels from the denser medium (A, $n=2$ ) to the rarer medium (B, $n=\sqrt{2}$ ). Using Snell's law at the critical angle gives $\theta_c = 45°$ .

When light travels from a denser medium to a rarer medium, there exists a special angle of incidence beyond which no refraction occurs—all light is reflected back. This is total internal reflection, and the boundary angle is the critical angle.

The critical angle exists only when light moves from higher refractive index to lower refractive index. Here, medium A has $n_A = 2$ and medium B has $n_B = \sqrt{2}$ . Since $2 > \sqrt{2}$ , light traveling from A to B can undergo total internal reflection.

At the critical angle $\theta_c$ , the refracted ray grazes along the interface, meaning the angle of refraction is $90°$ . Snell's law at the interface gives us:

n_A \sin\theta_c = n_B \sin 90°

Let me work through this step by step:

Apply Snell's law at the critical condition When the angle of incidence in medium A equals the critical angle, the refracted ray in medium B travels parallel to the interface:

n_A \sin\theta_c = n_B \times 1

Substitute the given refractive indices

2 \sin\theta_c = \sqrt{2}

Solve for $\sin\theta_c$ $\sin\theta_c = \frac{\sqrt{2}}{2}$ …

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Q.The refractive indices of two media A and B are 222 and 2\sqrt{2}2​ respectively. What is the critical angle for their interface ?

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Q.The refractive indices of two media A and B are $2$ and $\sqrt{2}$ respectively. What is the critical angle for their interface ?

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