If `theta` is the polarising angle for two optical media whose critical angles are `C_1 and C_2` , then the correct relation is

Options

A. `sintheta=(sinC_2)/(sinC_1)`
B. `theta=(sinC_2)/(sinC_1)`
C. `tantheta=(sinC_1)/(sinC_2)`
D. `sintheta=(sinC_1)/(sinC_2)`

Answer

To solve the problem, we need to establish the relationship between the polarizing angle (θ) and the critical angles (C1 and C2) for two optical media.

### Step-by-Step Solution:

1. **Understanding Brewster's Law**: According to Brewster's Law, the tangent of the polarizing angle (θ) is given by the ratio of the refractive indices of the two media:
\[ \tan \theta = \frac{n_2}{n_1} \]

2. **Relating Refractive Index to Critical Angle**: The critical angle (C) for total internal reflection can be expressed using Snell's Law. For a medium with refractive index \( n_1 \) (denser) and air (with refractive index \( n_2 = 1 \)), the critical angle \( C_1 \) can be defined as:
\[ n_1 \sin C_1 = n_2 \sin 90^\circ \]
Since \( \sin 90^\circ = 1 \), we can simplify this to:
\[ n_1 = \frac{1}{\sin C_1} \]

3. **Applying the Same for the Second Medium**: Similarly, for the second medium with critical angle \( C_2 \):
\[ n_2 = \frac{1}{\sin C_2} \]

4. **Substituting the Refractive Indices**: Now we can substitute the expressions for \( n_1 \) and \( n_2 \) into the equation from Brewster's Law:
\[ \tan \theta = \frac{n_2}{n_1} = \frac{\frac{1}{\sin C_2}}{\frac{1}{\sin C_1}} = \frac{\sin C_1}{\sin C_2} \]

5. **Final Relation**: Thus, we arrive at the relationship:
\[ \tan \theta = \frac{\sin C_1}{\sin C_2} \]

### Conclusion:
The correct relation between the polarizing angle (θ) and the critical angles (C1 and C2) is:
\[ \tan \theta = \frac{\sin C_1}{\sin C_2} \]