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/sqroot2` 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, we need to find the velocity of light in a medium given that the critical angle for total internal reflection (TIR) from the medium to vacuum is 45 degrees.

### Step-by-Step Solution:

1. **Understanding the Critical Angle**:
The critical angle (θ_c) is the angle of incidence above which total internal reflection occurs. The relationship between the critical angle and the refractive index (μ) of the medium is given by:
\[ \sin(\theta_c) = \frac{1}{\mu} \]

2. **Substituting the Given Value**:
We know that the critical angle θ_c is 45 degrees. Thus, we can substitute this value into the equation:
\[ \sin(45^\circ) = \frac{1}{\mu} \]

3. **Calculating the Sine of 45 Degrees**:
The sine of 45 degrees is:
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \]
Therefore, we can rewrite the equation as:
\[ \frac{1}{\sqrt{2}} = \frac{1}{\mu} \]

4. **Finding the Refractive Index (μ)**:
Rearranging the equation gives us:
\[ \mu = \sqrt{2} \]

5. **Relating Refractive Index to Velocity**:
The refractive index (μ) is also related to the speed of light in vacuum (C) and the speed of light in the medium (V) by the formula:
\[ \mu = \frac{C}{V} \]
Rearranging this gives:
\[ V = \frac{C}{\mu} \]

6. **Substituting the Values**:
We know that the speed of light in vacuum (C) is approximately \(3 \times 10^8\) m/s. Substituting the value of μ:
\[ V = \frac{3 \times 10^8 \text{ m/s}}{\sqrt{2}} \]

7. **Final Calculation**:
Thus, the speed of light in the medium is:
\[ V = \frac{3}{\sqrt{2}} \times 10^8 \text{ m/s} \]

### Final Answer:
The velocity of light in the medium is:
\[ V = \frac{3}{\sqrt{2}} \times 10^8 \text{ m/s} \]

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