A ray of light is incident on the surface of separation of a medium at an angle `45^(@)` and is refracted in the medium at an angle `30^(@)`. What will be the velocity of light in the medium?

Options

A. `1.96 xx 10^(8)` m/s
B. `2.12 xx 10^(8)` m/s
C. `3.86 xx 10^(8)` m/s
D. `3.33 xx 18^(8)`m/s

Answer

To solve the problem of finding the velocity of light in the medium given the angles of incidence and refraction, we can follow these steps:

### Step 1: Identify the known values
- Angle of incidence (i) = 45 degrees
- Angle of refraction (r) = 30 degrees
- Refractive index of air (µ1) = 1 (since it is air)

### Step 2: Apply Snell's Law
Snell's Law states that:
\[ \mu_1 \sin(i) = \mu_2 \sin(r) \]
Where:
- \( \mu_1 \) = refractive index of the first medium (air)
- \( \mu_2 \) = refractive index of the second medium (unknown)
- \( i \) = angle of incidence
- \( r \) = angle of refraction

Substituting the known values into Snell's Law:
\[ 1 \cdot \sin(45^\circ) = \mu_2 \cdot \sin(30^\circ) \]

### Step 3: Calculate the sine values
We know:
- \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

Substituting these values into the equation:
\[ 1 \cdot \frac{1}{\sqrt{2}} = \mu_2 \cdot \frac{1}{2} \]

### Step 4: Solve for the refractive index (µ2)
Rearranging the equation to find \( \mu_2 \):
\[ \mu_2 = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} \]
\[ \mu_2 = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 5: Relate the refractive index to the velocity of light
The refractive index is also defined as:
\[ \mu = \frac{c}{v} \]
Where:
- \( c \) = speed of light in vacuum (approximately \( 3 \times 10^8 \) m/s)
- \( v \) = speed of light in the medium

Rearranging this gives:
\[ v = \frac{c}{\mu} \]

### Step 6: Substitute the values to find v
Substituting \( \mu = \sqrt{2} \):
\[ v = \frac{3 \times 10^8}{\sqrt{2}} \]

### Step 7: Calculate the numerical value
Calculating \( v \):
\[ v \approx \frac{3 \times 10^8}{1.414} \approx 2.12 \times 10^8 \text{ m/s} \]

### Final Answer
The velocity of light in the medium is approximately \( 2.12 \times 10^8 \) m/s.

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