When a light ray is incident on a medium of refractive index 4/3, reflected light is found to be 100% polarized. What is the angle of refraction in degrees?
`(tan 53^@ = 4/3)`.

Explanation

To solve the problem, we can follow these steps:

### Step 1: Understand the concept of Brewster's Angle
When light is incident on a medium and the reflected light is 100% polarized, it occurs at Brewster's angle (θ_B). At this angle, the reflected and refracted rays are perpendicular to each other.

### Step 2: Set up the relationship between angles
At Brewster's angle, we have:
\[ I + R = 90^\circ \]
where \(I\) is the angle of incidence and \(R\) is the angle of refraction. Therefore, we can express the angle of incidence as:
\[ I = 90^\circ - R \]

### Step 3: Use Snell's Law
According to Snell's Law:
\[ \mu_1 \sin I = \mu_2 \sin R \]
where \(\mu_1\) is the refractive index of air (approximately 1) and \(\mu_2\) is the refractive index of the medium (given as \(4/3\)). Thus, we can write:
\[ 1 \cdot \sin I = \frac{4}{3} \sin R \]

### Step 4: Substitute for \(I\)
Substituting \(I = 90^\circ - R\) into Snell's Law gives:
\[ \sin(90^\circ - R) = \frac{4}{3} \sin R \]
Using the identity \(\sin(90^\circ - R) = \cos R\), we have:
\[ \cos R = \frac{4}{3} \sin R \]

### Step 5: Rearranging the equation
Rearranging the equation gives:
\[ \frac{\cos R}{\sin R} = \frac{4}{3} \]
This can be rewritten as:
\[ \cot R = \frac{4}{3} \]

### Step 6: Finding the angle \(R\)
To find \(R\), we can take the cotangent inverse:
\[ R = \cot^{-1}\left(\frac{4}{3}\right) \]
Using the known value, we find:
\[ R \approx 37^\circ \]

### Final Answer
The angle of refraction \(R\) is approximately \(37^\circ\).

---