A light ray incidents on water `(n=4/3)` surface from air and reflected from air and reflected part of it is found to be polarized. Find the deviation of refracted light from its original path:

Options

A. `53^(@)`
B. `90^(@)`
C. `74^(@)`
D. `16^(@)`

Explanation

To solve the problem, we need to find the deviation of the refracted light from its original path when a light ray strikes the surface of water from air and part of it is polarized. Here’s a step-by-step solution:

### Step 1: Understand the Situation
When a light ray strikes the interface between air and water, part of the light is reflected and part is refracted. The reflected light is found to be polarized, which indicates that the angle of incidence is equal to the angle of polarization.

### Step 2: Use the Angle of Polarization
The angle of polarization \( I_p \) can be calculated using the formula:
\[ I_p = \tan^{-1}(\mu) \]
where \( \mu \) is the refractive index of water. Given that the refractive index of water is \( \frac{4}{3} \):
\[ I_p = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 3: Calculate the Angle of Polarization
Using a calculator or trigonometric tables, we find:
\[ I_p \approx 53^\circ \]

### Step 4: Apply Snell's Law
According to Snell's law, the relationship between the angles of incidence \( I \) and refraction \( R \) is given by:
\[ \sin I = n \cdot \sin R \]
where \( n \) is the refractive index of the second medium (water in this case).

### Step 5: Calculate the Deviation
The deviation \( \Delta \) caused by the refraction can be calculated using:
\[ \Delta = I - R \]
We know that \( R \) can be expressed in terms of \( I \):
\[ R = \sin^{-1}\left(\frac{\sin I}{n}\right) \]
Substituting \( n = \frac{4}{3} \) and \( I = 53^\circ \):
\[ R = \sin^{-1}\left(\frac{\sin(53^\circ)}{\frac{4}{3}}\right) \]

### Step 6: Calculate \( R \)
First, calculate \( \sin(53^\circ) \):
\[ \sin(53^\circ) \approx 0.7986 \]
Now, substituting into the equation for \( R \):
\[ R = \sin^{-1}\left(\frac{0.7986}{\frac{4}{3}}\right) = \sin^{-1}\left(0.59895\right) \]
Calculating \( R \):
\[ R \approx 37^\circ \]

### Step 7: Find the Deviation
Now substituting \( I \) and \( R \) back into the deviation formula:
\[ \Delta = 53^\circ - 37^\circ = 16^\circ \]

### Final Answer
The deviation of the refracted light from its original path is:
\[ \Delta = 16^\circ \]