If a point source is placed at a depth `h` in a liquid of refractive index `4/3` . Find percentage of energy of light that escapes from liquid. (assuming `100%` transmission of emerging light)

Options

A. `15%`
B. `17%`
C. `21%`
D. `34%`

Answer

To solve the problem of finding the percentage of energy of light that escapes from a liquid with a refractive index of \( \frac{4}{3} \), we will follow these steps:

### Step 1: Understanding the Problem
We have a point source of light placed at a depth \( h \) in a liquid with a refractive index \( n = \frac{4}{3} \). We need to determine the percentage of light energy that escapes from the liquid.

### Step 2: Apply Snell's Law
Using Snell's Law, we can find the angle of incidence \( i \) at which light emerges from the liquid to air. Snell's law states:
\[ n_1 \sin i = n_2 \sin r \]
Here, \( n_1 = \frac{4}{3} \) (for the liquid), \( n_2 = 1 \) (for air), and \( r \) is the angle of refraction. Since we are interested in the critical angle for total internal reflection (TIR), we can set \( r = 90^\circ \):
\[ \frac{4}{3} \sin i = 1 \cdot \sin 90^\circ \]
This simplifies to:
\[ \sin i = \frac{3}{4} \]

### Step 3: Calculate \( \cos i \)
Using the identity \( \sin^2 i + \cos^2 i = 1 \):
\[ \cos i = \sqrt{1 - \sin^2 i} = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]

### Step 4: Determine the Solid Angle
The solid angle \( \Omega \) subtended at the apex of the point source can be calculated using the formula:
\[ \Omega = 2\pi(1 - \cos \beta) \]
where \( \beta \) is the angle of emergence. In our case, \( \cos \beta = \cos i = \frac{\sqrt{7}}{4} \):
\[ \Omega = 2\pi \left(1 - \frac{\sqrt{7}}{4}\right) \]

### Step 5: Calculate the Total Solid Angle
The total solid angle in three-dimensional space is \( 4\pi \). The percentage of light escaping can be calculated by taking the ratio of the solid angle that escapes to the total solid angle:
\[ \text{Percentage of light escaping} = \frac{\Omega}{4\pi} \times 100 = \frac{2\pi \left(1 - \frac{\sqrt{7}}{4}\right)}{4\pi} \times 100 \]
This simplifies to:
\[ \text{Percentage of light escaping} = \frac{1 - \frac{\sqrt{7}}{4}}{2} \times 100 \]

### Step 6: Final Calculation
Substituting \( \sqrt{7} \approx 2.64575 \):
\[ \text{Percentage of light escaping} = \frac{1 - \frac{2.64575}{4}}{2} \times 100 \approx \frac{1 - 0.6614375}{2} \times 100 \approx \frac{0.3385625}{2} \times 100 \approx 16.93\% \]
Rounding this gives approximately \( 17\% \).

### Conclusion
The percentage of energy of light that escapes from the liquid is approximately \( 17\% \).

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