Let the refractive index of a denser medium with respect to a rarer medium be `n_(12)` and its critical angle be `theta_(c)`. At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is `90^(@)`, Angle A given by -

Options

A. `tan^(-1)(sin theta_(c))`
B. `(1)/(tan^(-1)(sintheta_(c)))`
C. `cos^(-1)(sin theta_(C))`
D. `(1)/(cos^(-1)(sintheta_(c)))`

Answer

To solve the problem, we need to find the angle of incidence \( A \) when light travels from a denser medium to a rarer medium, given that the angle between the reflected and refracted rays is \( 90^\circ \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let \( n_{12} \) be the refractive index of the denser medium with respect to the rarer medium.
- The critical angle is denoted as \( \theta_c \).
- When light travels from the denser medium to the rarer medium, part of the light is reflected, and part is refracted. The angle between the reflected ray and the refracted ray is \( 90^\circ \).

2. **Applying the Law of Reflection**:
- According to the law of reflection, the angle of incidence \( A \) is equal to the angle of reflection. Therefore, if the angle of incidence is \( A \), the angle of reflection is also \( A \).

3. **Determining the Angle of Refraction**:
- Since the angle between the reflected and refracted rays is \( 90^\circ \), the angle of refraction \( r \) can be expressed as:
\[ r = 90^\circ - A \]

4. **Using Snell's Law**:
- Snell's law states that:
\[ n_{12} \sin A = n_{21} \sin r \]
- Here, \( n_{21} \) is the refractive index of the rarer medium with respect to the denser medium, which is \( \frac{1}{n_{12}} \).
- Therefore, we can rewrite Snell's law as:
\[ n_{12} \sin A = \frac{1}{n_{12}} \sin(90^\circ - A) \]

5. **Simplifying the Equation**:
- Since \( \sin(90^\circ - A) = \cos A \), we can substitute this into the equation:
\[ n_{12} \sin A = \frac{1}{n_{12}} \cos A \]

6. **Rearranging the Equation**:
- Multiplying both sides by \( n_{12} \):
\[ n_{12}^2 \sin A = \cos A \]

7. **Dividing Both Sides by \( \cos A \)**:
- We can rearrange this to find \( \tan A \):
\[ n_{12}^2 \tan A = 1 \]
- Therefore:
\[ \tan A = \frac{1}{n_{12}^2} \]

8. **Finding the Angle \( A \)**:
- Finally, we can express \( A \) as:
\[ A = \tan^{-1}\left(\frac{1}{n_{12}^2}\right) \]

### Final Answer:
\[ A = \tan^{-1}\left(\frac{1}{n_{12}^2}\right) \]