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 find the fraction of light energy that escapes directly from the liquid surface when a point source of light is placed at a depth \( h = 0.5 \, \text{m} \) below the surface of a liquid with a 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 the formula:
\[ \sin \theta_c = \frac{1}{\mu} \]
Substituting the given 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 \( \cos \theta_c \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \cos \theta_c = \sqrt{1 - \sin^2 \theta_c} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 3: Calculate the Solid Angle Subtended by the Cone
The solid angle \( \Omega \) subtended by the cone at the boundary can be calculated using:
\[ \Omega = 2\pi(1 - \cos \theta_c) \]
Substituting \( \cos \theta_c \):
\[ \Omega = 2\pi\left(1 - \frac{3}{5}\right) = 2\pi\left(\frac{2}{5}\right) = \frac{4\pi}{5} \]

### Step 4: Calculate the Total Solid Angle of a Sphere
The total solid angle of a sphere is:
\[ 4\pi \]

### Step 5: Calculate the Fraction of Light Energy Lost
The fraction of light energy lost \( E_L \) can be expressed as:
\[ E_L = \frac{\Omega}{4\pi} \]
Substituting the values:
\[ E_L = \frac{\frac{4\pi}{5}}{4\pi} = \frac{4}{5} \cdot \frac{1}{4} = \frac{1}{5} = 0.2 \]

### Final Result
The fraction of light energy that escapes directly from the liquid surface is:
\[ E_L = 0.2 \]