Question

A rectangular glass slab ABCD of refractive index \(n_1\) is placed in a liquid of refractive index \(n_2\) (\(n_1 > n_2\)). A ray of light is incident at the surface AB of the slab as shown in the figure. Find out the maximum value of angle \(\alpha\), so that the ray comes out only from the other surface CD.

Hint:

Apply Snell's law at AB and impose total internal reflection at the side face, where the internal angle is \((90^\circ-r)\) and \(\sin\theta_c=n_2/n_1\).

Answer 1

Step 1: Understand the required condition.
The ray enters face AB, bends inside the glass and travels towards the top (or bottom) side face. For the ray to leave only from the far face CD, it must undergo total internal reflection (TIR) at the side face it meets, instead of refracting out of it. The largest \(\alpha\) corresponds to the ray striking that side face exactly at the critical angle.

Step 2: Apply Snell's law at face AB.
Light goes from the liquid (\(n_2\)) into the glass (\(n_1\)). If \(r\) is the angle of refraction inside the glass,
\[ n_2 \sin\alpha = n_1 \sin r \quad\Rightarrow\quad \sin r = \frac{n_2}{n_1}\sin\alpha \qquad (1) \]
Step 3: Geometry at the side face.
The refracted ray then meets a side face whose normal is perpendicular to the normal of AB. So the angle of incidence at that face is \((90^\circ - r)\).

Step 4: Critical-angle (TIR) condition at the side face.
The critical angle \(\theta_c\) for the glass-liquid boundary satisfies \(\sin\theta_c = \dfrac{n_2}{n_1}\). For the maximum \(\alpha\), the ray strikes the side face exactly at \(\theta_c\):
\[ 90^\circ - r = \theta_c \quad\Rightarrow\quad \sin(90^\circ - r) = \sin\theta_c \]\[ \cos r = \frac{n_2}{n_1} \qquad (2) \]
Step 5: Find \(\sin r\) from equation (2).
\[ \sin r = \sqrt{1-\cos^2 r} = \sqrt{1-\frac{n_2^{\,2}}{n_1^{\,2}}} = \frac{\sqrt{n_1^{\,2}-n_2^{\,2}}}{n_1} \qquad (3) \]
Step 6: Combine (1) and (3).
\[ \frac{n_2}{n_1}\sin\alpha_{max} = \frac{\sqrt{n_1^{\,2}-n_2^{\,2}}}{n_1} \]\[ \sin\alpha_{max} = \frac{\sqrt{n_1^{\,2}-n_2^{\,2}}}{n_2} \]
Step 7: Write the final answer.
\[\boxed{\ \alpha_{max} = \sin^{-1}\!\left(\frac{\sqrt{n_1^{\,2}-n_2^{\,2}}}{n_2}\right)\ }\]For any \(\alpha\) up to this value, the ray totally reflects at the side face and emerges only from CD.