If critical angle for TIR from medium to vacuum is 45 deg then velocity of light in medium

Options

A. `1.5 x 10^8 m/s`
B. `3/root 2 x 10^8 m/s`
C. `root 2 x 10^8 m/s`
D. `3 x 10^8 m/s`

Answer

To solve the problem of finding the velocity of light in a medium where the critical angle for total internal reflection (TIR) from that medium to vacuum is 45 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Critical Angle**:
The critical angle (C) is the angle of incidence above which total internal reflection occurs. For a medium to vacuum, the critical angle can be defined using Snell's law:
\[ \mu_1 \sin C = \mu_2 \sin 90^\circ \]
Here, \(\mu_1\) is the refractive index of the medium, and \(\mu_2\) is the refractive index of vacuum, which is 1.

2. **Setting Up the Equation**:
Since \(\sin 90^\circ = 1\), we can simplify the equation to:
\[ \mu_1 \sin C = 1 \]
Rearranging gives:
\[ \sin C = \frac{1}{\mu_1} \]

3. **Substituting the Critical Angle**:
Given that the critical angle \(C\) is 45 degrees, we can substitute this value into the equation:
\[ \sin 45^\circ = \frac{1}{\mu_1} \]
We know that \(\sin 45^\circ = \frac{1}{\sqrt{2}}\), so:
\[ \frac{1}{\sqrt{2}} = \frac{1}{\mu_1} \]

4. **Finding the Refractive Index**:
From the equation above, we can find \(\mu_1\):
\[ \mu_1 = \sqrt{2} \]

5. **Relating Velocity and Refractive Index**:
The velocity of light in a medium (\(v\)) is related to the speed of light in vacuum (\(c\)) and the refractive index (\(\mu\)) by the formula:
\[ v = \frac{c}{\mu} \]
Substituting \(\mu_1\) into this equation:
\[ v = \frac{c}{\sqrt{2}} \]

6. **Substituting the Value of \(c\)**:
The speed of light in vacuum is approximately \(c = 3 \times 10^8 \, \text{m/s}\). Thus:
\[ v = \frac{3 \times 10^8}{\sqrt{2}} \, \text{m/s} \]

7. **Calculating the Final Value**:
To find the numerical value:
\[ v \approx \frac{3 \times 10^8}{1.414} \approx 2.121 \times 10^8 \, \text{m/s} \]

### Final Answer:
The velocity of light in the medium is approximately:
\[ v \approx 2.121 \times 10^8 \, \text{m/s} \]