A point source of light S is placed at the bottom of a vessel containg a liquid of refractive index `5//3.` A person is viewing the source from above the surface. There is an opaque disc of radius `1 cm` floating on the surface. The centre of the disc lies vertically above the source S. The liquid from the vessel is gradually drained out through a tap. What is the maximum height of the liquid for which the source cannot at all be seen from above?

Explanation

To solve the problem step by step, we will analyze the situation involving the point source of light, the opaque disc, and the refractive index of the liquid.

### Step 1: Understand the Setup
- We have a point source of light \( S \) at the bottom of a vessel filled with a liquid of refractive index \( \mu = \frac{5}{3} \).
- An opaque disc of radius \( r = 1 \, \text{cm} \) is floating on the surface of the liquid, directly above the source \( S \).
- We need to determine the maximum height \( h \) of the liquid such that the source \( S \) cannot be seen from above the surface.

### Step 2: Identify the Critical Angle
- The critical angle \( I_c \) can be calculated using the formula:
\[ \mu = \frac{1}{\sin I_c} \]
- Rearranging gives:
\[ \sin I_c = \frac{1}{\mu} = \frac{3}{5} \]
- Therefore, the critical angle \( I_c \) can be calculated as:
\[ I_c = \sin^{-1}\left(\frac{3}{5}\right) \]

### Step 3: Relate the Geometry
- When the light travels from the liquid to the air, the angle of incidence \( I \) at the surface of the liquid must be equal to the critical angle \( I_c \) for the light to be totally internally reflected.
- The geometry of the situation gives us:
- The radius \( r \) of the disc is the horizontal distance from the center of the disc to the edge.
- The height \( h \) of the liquid is the vertical distance from the surface of the liquid to the source \( S \).

### Step 4: Use Trigonometry
- From the right triangle formed by the height \( h \) and radius \( r \), we can use the relationship:
\[ \tan I_c = \frac{r}{h} \]
- Rearranging gives:
\[ h = \frac{r}{\tan I_c} \]

### Step 5: Calculate \( \tan I_c \)
- We know:
\[ \sin I_c = \frac{3}{5} \quad \text{and} \quad \cos I_c = \sqrt{1 - \sin^2 I_c} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
- Therefore, we can find \( \tan I_c \):
\[ \tan I_c = \frac{\sin I_c}{\cos I_c} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

### Step 6: Substitute Values
- Now substituting \( r = 1 \, \text{cm} \) and \( \tan I_c = \frac{3}{4} \) into the equation for \( h \):
\[ h = \frac{1 \, \text{cm}}{\frac{3}{4}} = \frac{4}{3} \, \text{cm} \]

### Conclusion
- The maximum height of the liquid for which the source cannot be seen from above is:
\[ h = \frac{4}{3} \, \text{cm} \]