The velocity of light in air is `3xx10^(8)ms^(-1)` and that in water is `2.2xx10^(8)ms^(-1).` Find the polarising angle of incidence.

Options

A. `45^(@)`
B. `50^(@)`
C. `53.74^(@)`
D. `63^(@)`

Explanation

To find the polarizing angle of incidence when light travels from air into water, we can follow these steps:

### Step 1: Identify the given values
- Velocity of light in air, \( C = 3 \times 10^8 \, \text{m/s} \)
- Velocity of light in water, \( V = 2.2 \times 10^8 \, \text{m/s} \)

### Step 2: Calculate the refractive index \( n \)
The refractive index \( n \) of a medium is given by the formula:
\[ n = \frac{C}{V} \]
Substituting the values:
\[ n = \frac{3 \times 10^8}{2.2 \times 10^8} \]
The \( 10^8 \) cancels out:
\[ n = \frac{3}{2.2} \approx 1.36 \]

### Step 3: Apply Brewster's Law
Brewster's Law states that the refractive index \( n \) is equal to the tangent of the polarizing angle \( \theta_p \):
\[ n = \tan(\theta_p) \]
Substituting the value of \( n \):
\[ 1.36 = \tan(\theta_p) \]

### Step 4: Calculate the polarizing angle \( \theta_p \)
To find \( \theta_p \), we take the arctangent (inverse tangent) of \( 1.36 \):
\[ \theta_p = \tan^{-1}(1.36) \]
Using a calculator, we find:
\[ \theta_p \approx 53.74^\circ \]

### Final Answer
The polarizing angle of incidence is approximately \( 53.74^\circ \).

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