Question

A point source of light S is placed at the bottom of a vessel containing a liquid of refracting index $\frac{5}{3}$. A person is viewing the source from above the surface. there is an opaque disc of radius 2 cm floating on the surface of the liquid such that the center of the disc is vertically above the source S. The liquid from the vessel is gradually drained out through a tap. The maximum height of the liquid for which the source cannot be seen from above is

• A. 1.33 cm
• B. 2.66 cm
• C. 3.99 cm
• D. 0.33 m

Answer 1

Answer: (B)

The maximum height of the liquid is $\frac{8}{3}\,\text{cm}$ (≈2.66 cm)

Answer 2

1. The liquid’s refractive index is $\mu = \frac{5}{3}$. At the water-air interface the critical angle $\theta_c$ is given by

   $$\sin \theta_c = \frac{n_{\text{air}}}{n_{\text{water}}} = \frac{1}{5/3} = \frac{3}{5}.$$

2. The corresponding cosine and tangent are:

   $$\cos \theta_c = \sqrt{1-\sin^2 \theta_c} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}, \quad \tan \theta_c = \frac{\sin \theta_c}{\cos \theta_c} = \frac{3/5}{4/5} = \frac{3}{4}.$$

3. A point source $S$ is at the bottom and an opaque disc of radius $r = 2\,\text{cm}$ floats on the surface exactly above $S$. The light emerging from $S$ will be confined within a circle of radius

   $$R = h \tan \theta_c$$

   where $h$ is the depth of the liquid above $S$.

4. For an observer above the surface not to see $S$, the disc must cover the entire area from which light can emerge; hence:

   $$h \tan \theta_c \le 2\,\text{cm}.$$

5. Substituting $\tan \theta_c = \frac{3}{4}$:

   $$h \cdot \frac{3}{4} = 2 \quad \Longrightarrow \quad h = \frac{2 \times 4}{3} = \frac{8}{3}\,\text{cm} \approx 2.66\,\text{cm}.$$

Explanation (Core Minimal):

• Calculate critical angle using $\sin \theta_c = \frac{3}{5}$, thus $\tan \theta_c = \frac{3}{4}$.
• For the disc to hide the entire light emerging from $S$, set $h\tan \theta_c = 2\,\text{cm}$, so $h = \frac{8}{3}\,\text{cm} \approx 2.66\,\text{cm}$.