If a glass plate refractive index is 1.732 is to be used as a polariser, what would be the angle of refraction?

Options

A. `30^(@)`
B. `60^(@)`
C. `45^(@)`
D. `35^(@)`

Explanation

To solve the problem of finding the angle of refraction when a glass plate with a refractive index of 1.732 is used as a polarizer, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Brewster's Angle**:
Brewster's angle (\(I_P\)) is the angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. The relationship is given by:
\[ \tan(I_P) = \mu \]
where \(\mu\) is the refractive index of the glass.

2. **Calculate Brewster's Angle**:
Given that the refractive index (\(\mu\)) of the glass is 1.732, we can find Brewster's angle:
\[ I_P = \tan^{-1}(1.732) \]
Since \(1.732\) is approximately \(\sqrt{3}\), we know:
\[ I_P = 60^\circ \]

3. **Applying Snell's Law**:
Snell's Law relates the angles and indices of refraction:
\[ \mu_1 \sin(I_P) = \mu_2 \sin(R) \]
Here, \(\mu_1\) (the refractive index of air) is approximately 1, \(\mu_2\) (the refractive index of glass) is 1.732, \(I_P\) is the angle of incidence (60°), and \(R\) is the angle of refraction.

4. **Substituting Values**:
Substitute the known values into Snell's Law:
\[ 1 \cdot \sin(60^\circ) = 1.732 \cdot \sin(R) \]
We know \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), so:
\[ \frac{\sqrt{3}}{2} = 1.732 \cdot \sin(R) \]

5. **Solving for \(\sin(R)\)**:
Rearranging gives:
\[ \sin(R) = \frac{\frac{\sqrt{3}}{2}}{1.732} \]
Simplifying this:
\[ \sin(R) = \frac{\sqrt{3}}{2 \cdot 1.732} = \frac{\sqrt{3}}{3.464} = \frac{1}{2} \]

6. **Finding Angle of Refraction**:
Now, we find the angle \(R\) such that:
\[ R = \sin^{-1}\left(\frac{1}{2}\right) \]
This gives:
\[ R = 30^\circ \]

### Final Answer:
The angle of refraction \(R\) when the glass plate is used as a polarizer is \(30^\circ\).

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