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. `tan^(-1)(sin theta)`
B. `tan^(-1)(sin phi)`
C. `sin^(-1)(tan theta)`
D. `sin^(-1)(tanphi)`

Explanation

To find the angle of refraction when a ray of light traveling through air enters a transparent medium at an angle equal to its Brewster angle, we can follow these steps:

### Step 1: Understand the Critical Angle and Brewster's Angle
The critical angle (φ) is the angle of incidence in the denser medium (in this case, the transparent medium) beyond which light cannot pass into the less dense medium (air) and is instead totally internally reflected. Brewster's angle (θ) is the angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection.

### Step 2: Use Snell's Law at the Critical Angle
According to Snell's Law:
\[ n_1 \sin(\theta_i) = n_2 \sin(\theta_r) \]
where:
- \(n_1\) is the refractive index of air (approximately 1),
- \(\theta_i\) is the angle of incidence (which will be the critical angle φ),
- \(n_2\) is the refractive index of the transparent medium,
- \(\theta_r\) is the angle of refraction (which will be 90° at the critical angle).

Since at the critical angle, \(\theta_r\) is 90°, we have:
\[ n_1 \sin(\phi) = n_2 \sin(90°) \]
This simplifies to:
\[ \sin(\phi) = n_2 \]

### Step 3: Apply Snell's Law at Brewster's Angle
At Brewster's angle (θ), the angle of incidence is equal to θ, and the angle of refraction can be expressed as:
\[ n_1 \sin(\theta) = n_2 \sin(r) \]
Here, \(\theta\) is the Brewster angle, and \(r\) is the angle of refraction we want to find.

### Step 4: Relate Brewster's Angle to the Angles
At Brewster's angle, the refracted ray and the reflected ray are perpendicular to each other. Thus:
\[ r + \theta = 90° \]
This implies:
\[ r = 90° - \theta \]

### Step 5: Substitute into Snell's Law
Substituting \(r\) into Snell's Law gives:
\[ \sin(\theta) = n_2 \sin(90° - \theta) \]
Using the identity \(\sin(90° - x) = \cos(x)\), we get:
\[ \sin(\theta) = n_2 \cos(\theta) \]

### Step 6: Substitute for \(n_2\)
From Step 2, we know that \(n_2 = \sin(\phi)\). Thus:
\[ \sin(\theta) = \sin(\phi) \cos(\theta) \]

### Step 7: Solve for the Angle of Refraction
Rearranging gives:
\[ \sin(\theta) = \sin(\phi) \cos(\theta) \]
This can be rewritten as:
\[ \tan(\theta) = \sin(\phi) \]
Thus, the angle of refraction \(r\) can be expressed as:
\[ r = 90° - \tan^{-1}(\sin(\phi)) \]

### Final Answer
The angle of refraction \(r\) is:
\[ r = 90° - \tan^{-1}(\sin(\phi)) \]