A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is `theta_(ic)` and Brewster's angle of incidence is `theta_(iB)` such that `(sintheta_(ic))/(sintheta_(iB))= eta=1.28` . The relative refractive index of the two media is

Options

A. 0.4
B. 0.2
C. 0.9
D. 0.8

Answer

To solve the problem, we need to find the relative refractive index of two media given the relationship between the critical angle and Brewster's angle. Let's denote the refractive indices of the denser medium as \( n_1 \) and the rarer medium as \( n_2 \).

### Step-by-Step Solution:

1. **Understanding the Critical Angle**:
The critical angle \( \theta_{ic} \) for total internal reflection is given by Snell's law:
\[ n_1 \sin(\theta_{ic}) = n_2 \sin(90^\circ) \]
Since \( \sin(90^\circ) = 1 \), we can rewrite this as:
\[ \sin(\theta_{ic}) = \frac{n_2}{n_1} \tag{1} \]

2. **Understanding Brewster's Angle**:
Brewster's angle \( \theta_{iB} \) is given by:
\[ \tan(\theta_{iB}) = \frac{n_2}{n_1} \]
This can also be expressed in terms of sine and cosine:
\[ \sin(\theta_{iB}) = \frac{n_2}{\sqrt{n_1^2 + n_2^2}} \tag{2} \]

3. **Using the Given Relationship**:
We are given that:
\[ \frac{\sin(\theta_{ic})}{\sin(\theta_{iB})} = 1.28 \]
Substituting equations (1) and (2) into this relationship gives:
\[ \frac{\frac{n_2}{n_1}}{\frac{n_2}{\sqrt{n_1^2 + n_2^2}}} = 1.28 \]
Simplifying this, we have:
\[ \frac{n_2 \sqrt{n_1^2 + n_2^2}}{n_2 n_1} = 1.28 \]
Cancelling \( n_2 \) (assuming \( n_2 \neq 0 \)):
\[ \frac{\sqrt{n_1^2 + n_2^2}}{n_1} = 1.28 \]

4. **Squaring Both Sides**:
Squaring both sides results in:
\[ \frac{n_1^2 + n_2^2}{n_1^2} = 1.28^2 \]
Calculating \( 1.28^2 \):
\[ 1.28^2 = 1.6384 \]
Thus, we have:
\[ 1 + \frac{n_2^2}{n_1^2} = 1.6384 \]
Rearranging gives:
\[ \frac{n_2^2}{n_1^2} = 1.6384 - 1 = 0.6384 \]

5. **Finding the Ratio of Refractive Indices**:
Taking the square root of both sides:
\[ \frac{n_2}{n_1} = \sqrt{0.6384} \approx 0.798 \]

6. **Conclusion**:
The relative refractive index of the two media is:
\[ \frac{n_2}{n_1} \approx 0.8 \]

### Final Answer:
The relative refractive index of the two media is approximately \( 0.8 \).