A light is incident on a refractive medium. The light is partially reflected and partially refracted. If both the reflected and refracted rays are mutually perpendicular, then determine the angle of refraction. Refractive index of medium is 1.47.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have light incident on a refractive medium, and we know that the reflected and refracted rays are mutually perpendicular. We need to find the angle of refraction given the refractive index of the medium (μ2) is 1.47.

### Step 2: Identify the Polarizing Angle
When the reflected and refracted rays are perpendicular, the angle of incidence (i) is known as Brewster's angle (Ip). According to Brewster's law, we have:
\[ \tan(I_p) = \mu \]
where \( \mu \) is the refractive index of the medium.

### Step 3: Calculate Brewster's Angle
Given that the refractive index (μ) is 1.47, we can calculate the Brewster's angle (Ip):
\[ I_p = \tan^{-1}(μ) = \tan^{-1}(1.47) \]

Using a calculator:
\[ I_p \approx 55.7^\circ \]

### Step 4: Apply Snell's Law
Now, we will apply Snell's law to find the angle of refraction (r). Snell's law states:
\[ \mu_1 \sin(i) = \mu_2 \sin(r) \]
where:
- \( \mu_1 = 1 \) (refractive index of air)
- \( \mu_2 = 1.47 \)
- \( i = I_p = 55.7^\circ \)

Substituting the known values into Snell's law:
\[ 1 \cdot \sin(55.7^\circ) = 1.47 \cdot \sin(r) \]

### Step 5: Calculate sin(r)
First, we need to calculate \( \sin(55.7^\circ) \):
\[ \sin(55.7^\circ) \approx 0.819 \]

Now substituting this value into the equation:
\[ 0.819 = 1.47 \cdot \sin(r) \]

Now, solve for \( \sin(r) \):
\[ \sin(r) = \frac{0.819}{1.47} \]
\[ \sin(r) \approx 0.558 \]

### Step 6: Calculate the Angle of Refraction
Now we find the angle of refraction (r) by taking the inverse sine:
\[ r = \sin^{-1}(0.558) \]

Using a calculator:
\[ r \approx 34.1^\circ \]

### Final Answer
The angle of refraction is approximately \( 34.1^\circ \).

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