(a) State the two conditions under which total internal reflection occurs. (b) A transparent container contain

Question

Q. (a) State the two conditions under which total internal reflection occurs.

(b) A transparent container contains layers of three immiscible transparent liquids A, B and C of refractive indices

n

\dfrac{3}{4}n

and

\dfrac{2}{3}n

, respectively. A laser beam is incident at the interface between A and B at an angle

\theta

as shown in figure. Prove that the beam does not enter region C at all for

\sin\theta \ge \dfrac{2}{3}

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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 beam fails to enter region C because it undergoes total internal reflection at the B–C interface. By applying Snell’s law at A–B and then the critical angle condition at B–C, we find that this happens precisely when $\sin\theta \ge \frac{2}{3}$ .

(a) Two conditions for total internal reflection

Light must travel from an optically denser medium to a rarer medium (i.e., $n_1 > n_2$ ).
The angle of incidence in the denser medium must exceed the critical angle for that pair of media.

(b) Proving the beam does not enter region C for $\sin\theta > \frac{2}{3}$

The key insight is that the beam must cross two interfaces in sequence: first from A to B, then from B to C. If total internal reflection occurs at the second interface (B–C), the beam never enters C. So we need to find the condition on $\theta$ that forces the angle of incidence at B–C to be at least the critical angle for that boundary.

Step 1: Refraction at the A–B interface

The beam starts in layer A (refractive index $n$ ) and enters layer B (refractive index $\frac{3}{4}n$ ). Let the angle of refraction in B be $r$ . By Snell’s law:

n \sin\theta = \frac{3}{4}n \sin r

Cancel $n$ (non-zero):

\sin\theta = \frac{3}{4} \sin r \quad\Rightarrow\quad \sin r = \frac{4}{3}\sin\theta

Note

Since the interfaces are parallel, the angle $r$ in B is also the angle of incidence at the B–C interface. This is a standard geometry trick — the refracted ray in B meets the next boundary at the same angle it made with the normal in B.

Step 2: Critical angle for the B–C interface

Layer B (index $\frac{3}{4}n$ ) is denser than layer C (index $\frac{2}{3}n$ ). The critical angle $\theta_c$ for the B–C boundary satisfies:

\sin\theta_c = \frac{n_C}{n_B} = \frac{\frac{2}{3}n}{\frac{3}{4}n} = \frac{2}{3} \times \frac{4}{3} = \frac{8}{9}

\sin\theta_c = \frac{8}{9}

Step 3: Condition for total internal reflection at B–C

For the beam to not enter C, the angle of incidence at B–C (which is $r$ ) must be at least the critical angle:

\sin r \ge \sin\theta_c

Substitute $\sin r = \frac{4}{3}\sin\theta$ and $\sin\theta_c = \frac{8}{9}$ :

$\frac{4}{3}\sin\theta \ge \frac{8}{9}$ …

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Q.(a) State the two conditions under which total internal reflection occurs.

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