The critical angle between a certain transparent medium and air is `phi`. A ray of light travelling through air enters the medium at an angle of incidence equal to its Brewster angle `theta.` Therefore, the angle of refraction is

Options

A. (a)`tan^(-1)(sin theta)`
B. (b)`tan^(-1)(sin phi)`
C. (c)`sin^(-1)(tan theta)`
D. (d)`sin^(-1)(tan phi)`

Explanation

To solve the problem, we need to find the angle of refraction when a ray of light traveling through air enters a certain transparent medium at an angle of incidence equal to its Brewster angle (θ). The critical angle between the medium and air is given as φ.

### Step-by-Step Solution:

1. **Understanding Brewster's Law**:
According to Brewster's law, the refractive index (μ) of a medium can be expressed in terms of the Brewster angle (θ):
\[ \mu = \tan(\theta) \]

2. **Relating Refractive Index to Critical Angle**:
The critical angle (φ) is related to the refractive index by the formula:
\[ \mu = \frac{1}{\sin(\phi)} \]

3. **Equating the Two Expressions for Refractive Index**:
Since both expressions represent the refractive index (μ), we can set them equal to each other:
\[ \tan(\theta) = \frac{1}{\sin(\phi)} \]

4. **Finding the Relationship Between θ and φ**:
Rearranging the equation gives us:
\[ \tan(\theta) \cdot \sin(\phi) = 1 \]

5. **Using the Angle of Refraction**:
When light travels from air to the medium, we can use Snell's law:
\[ n_1 \sin(\theta) = n_2 \sin(r) \]
where \( n_1 = 1 \) (for air) and \( n_2 = \mu \) (for the medium). Thus, we have:
\[ \sin(\theta) = \mu \sin(r) \]

6. **Substituting for μ**:
From step 3, we know that \( \mu = \tan(\theta) \). Therefore, we can write:
\[ \sin(\theta) = \tan(\theta) \sin(r) \]

7. **Using the Identity for Tangent**:
Recall that \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). Substituting this into the equation gives:
\[ \sin(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \sin(r) \]

8. **Solving for sin(r)**:
Dividing both sides by \( \sin(\theta) \) (assuming \( \sin(\theta) \neq 0 \)):
\[ 1 = \frac{\sin(r)}{\cos(\theta)} \]
Thus, we find:
\[ \sin(r) = \cos(\theta) \]

9. **Finding the Angle of Refraction (r)**:
Now, we can find the angle of refraction:
\[ r = \sin^{-1}(\cos(\theta)) \]

10. **Using the Complementary Angle Identity**:
Since \( \cos(\theta) = \sin(90^\circ - \theta) \), we can express \( r \) as:
\[ r = 90^\circ - \theta \]

11. **Final Expression for Angle of Refraction**:
From the earlier steps, we can also relate \( \theta \) to φ:
\[ \theta = 90^\circ - \phi \]
Therefore, substituting this back, we find:
\[ r = \phi \]

### Conclusion:
The angle of refraction \( r \) is equal to the critical angle \( \phi \).