A point source of light is placed at a depth h=0.5 m below the surface of a liquid `(mu=5/4)` , Then , the fraction of light energy that escape directly from the liquid surface is

Options

A. 0.1
B. 0.2
C. 0.3
D. 0.4

Answer

To solve the problem of finding the fraction of light energy that escapes directly from the surface of a liquid when a point source of light is placed at a depth \( h = 0.5 \, \text{m} \) below the surface of a liquid with refractive index \( \mu = \frac{5}{4} \), we can follow these steps:

### Step 1: Calculate the Critical Angle
The critical angle \( \theta_c \) can be calculated using Snell's law. When light travels from a denser medium (liquid) to a rarer medium (air), the critical angle is given by:

\[ \sin \theta_c = \frac{1}{\mu} \]

Substituting the value of \( \mu \):

\[ \sin \theta_c = \frac{1}{\frac{5}{4}} = \frac{4}{5} \]

Now, we find \( \theta_c \):

\[ \theta_c = \sin^{-1}\left(\frac{4}{5}\right) \]

### Step 2: Calculate the Solid Angle
The solid angle \( \omega \) subtended by the cone of light that escapes is given by:

\[ \omega = 2\pi(1 - \cos \theta_c) \]

To find \( \cos \theta_c \), we can use the identity:

\[ \cos \theta_c = \sqrt{1 - \sin^2 \theta_c} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

Substituting this into the solid angle formula:

\[ \omega = 2\pi\left(1 - \frac{3}{5}\right) = 2\pi\left(\frac{2}{5}\right) = \frac{4\pi}{5} \]

### Step 3: Calculate the Total Solid Angle
The total solid angle for a sphere is \( 4\pi \).

### Step 4: Calculate the Fraction of Light Energy Escaping
The fraction of light energy that escapes is the ratio of the solid angle that escapes to the total solid angle:

\[ \text{Fraction} = \frac{\omega}{4\pi} = \frac{\frac{4\pi}{5}}{4\pi} = \frac{1}{5} \]

### Conclusion
Thus, the fraction of light energy that escapes directly from the liquid surface is:

\[ \text{Fraction} = 0.2 \]