Critical angle for a certain wavelength of light in glass is `40^(@)`. Calculate the polarizing angle and the angle of refraction in glass corresponding to it.

Answer

To solve the problem, we need to find the polarizing angle and the angle of refraction in glass given that the critical angle for a certain wavelength of light in glass is \( 40^\circ \).

### Step 1: Calculate the refractive index of glass
The critical angle (\( C \)) is given by the formula:
\[ \mu \sin C = 1 \]
where \( \mu \) is the refractive index of glass. Given \( C = 40^\circ \):
\[ \mu \sin 40^\circ = 1 \]
Thus, we can find \( \mu \):
\[ \mu = \frac{1}{\sin 40^\circ} \]

### Step 2: Calculate \( \sin 40^\circ \)
Using a calculator or trigonometric tables, we find:
\[ \sin 40^\circ \approx 0.6428 \]
Now substituting this value into the equation for \( \mu \):
\[ \mu = \frac{1}{0.6428} \approx 1.555 \]

### Step 3: Find the polarizing angle (\( i_p \))
The polarizing angle is related to the refractive index by Brewster's Law:
\[ \tan i_p = \mu \]
Substituting the value of \( \mu \):
\[ \tan i_p = 1.555 \]
Now, we can find \( i_p \):
\[ i_p = \tan^{-1}(1.555) \]
Calculating this gives:
\[ i_p \approx 57.3^\circ \]

### Step 4: Find the angle of refraction (\( r \))
Using Snell's Law at the polarizing angle:
\[ \mu_1 \sin i_p = \mu_2 \sin r \]
Assuming \( \mu_1 = 1 \) (for air) and \( \mu_2 = 1.555 \) (for glass):
\[ \sin i_p = \mu \sin r \]
Substituting \( \mu \) and \( i_p \):
\[ \sin 57.3^\circ = 1.555 \sin r \]
Calculating \( \sin 57.3^\circ \):
\[ \sin 57.3^\circ \approx 0.8387 \]
Thus:
\[ 0.8387 = 1.555 \sin r \]
Solving for \( \sin r \):
\[ \sin r = \frac{0.8387}{1.555} \approx 0.539 \]
Now, finding \( r \):
\[ r = \sin^{-1}(0.539) \approx 32.7^\circ \]

### Final Results
- The polarizing angle \( i_p \approx 57.3^\circ \)
- The angle of refraction \( r \approx 32.7^\circ \)