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) ("sin"theta_(c)))`
C. `"cos"^(-1) ("sin"theta_(c))`
D. `(1)/("cos"^(-1)("sin"theta_(c)))`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Critical Angle**:
The critical angle \( \theta_c \) is defined as the angle of incidence in the denser medium at which the angle of refraction in the rarer medium is \( 90^\circ \). According to Snell's law:
\[ n_{12} \sin(\theta_c) = n_{21} \sin(90^\circ) \]
Since \( \sin(90^\circ) = 1 \), we can simplify this to:
\[ n_{12} \sin(\theta_c) = n_{21} \]

2. **Using Snell's Law**:
When light is incident at an angle \( A \) from the denser medium to the rarer medium, we can apply Snell's law:
\[ n_{12} \sin(A) = n_{21} \sin(R) \]
where \( R \) is the angle of refraction.

3. **Relationship Between Reflected and Refracted Rays**:
Given that the angle between the reflected ray and the refracted ray is \( 90^\circ \), we have:
\[ R + R' = 90^\circ \]
where \( R' \) is the angle of reflection. Since the angle of reflection is equal to the angle of incidence, we have:
\[ R' = A \quad \text{and thus} \quad R + A = 90^\circ \quad \Rightarrow \quad R = 90^\circ - A \]

4. **Substituting for R in Snell's Law**:
Substitute \( R \) into Snell's law:
\[ n_{12} \sin(A) = n_{21} \sin(90^\circ - A) \]
Using the identity \( \sin(90^\circ - A) = \cos(A) \), we can rewrite it as:
\[ n_{12} \sin(A) = n_{21} \cos(A) \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{\sin(A)}{\cos(A)} = \frac{n_{21}}{n_{12}} \quad \Rightarrow \quad \tan(A) = \frac{n_{21}}{n_{12}} \]

6. **Expressing in Terms of the Critical Angle**:
From the critical angle relationship, we know:
\[ \frac{n_{21}}{n_{12}} = \sin(\theta_c) \]
Thus, we can substitute this into our equation:
\[ \tan(A) = \sin(\theta_c) \]

7. **Finding A**:
Finally, we take the inverse tangent to find \( A \):
\[ A = \tan^{-1}(\sin(\theta_c)) \]

### Final Answer:
\[ A = \tan^{-1}(\sin(\theta_c)) \]