Light wave enters from medium 1 to medium 2. Its velocity in `2^(nd)` medium is double from `1^(st)`. For total internal reflection the angle of incidence must be greater than

Options

A. `30^@`
B. `60^@`
C. `45^@`
D. `90^@`

Answer

To solve the problem, we need to determine the critical angle for total internal reflection when light transitions from medium 1 to medium 2, where the velocity of light in medium 2 is double that in medium 1.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two media: medium 1 (with velocity \( V_1 \)) and medium 2 (with velocity \( V_2 \)).
- It is given that \( V_2 = 2V_1 \).

2. **Using the Relationship of Refractive Index**:
- The refractive index \( \mu \) of a medium is defined as:
\[ \mu = \frac{c}{V} \]
where \( c \) is the speed of light in vacuum and \( V \) is the speed of light in the medium.
- Therefore, for medium 1:
\[ \mu_1 = \frac{c}{V_1} \]
- For medium 2:
\[ \mu_2 = \frac{c}{V_2} = \frac{c}{2V_1} = \frac{1}{2} \cdot \frac{c}{V_1} = \frac{\mu_1}{2} \]

3. **Applying Snell's Law**:
- According to Snell's law:
\[ \mu_1 \sin i_c = \mu_2 \sin r \]
- For total internal reflection, the angle of refraction \( r \) is \( 90^\circ \). Thus, \( \sin r = 1 \).
- Therefore, we can write:
\[ \mu_1 \sin i_c = \mu_2 \]

4. **Substituting the Values**:
- Substitute \( \mu_2 = \frac{\mu_1}{2} \) into the equation:
\[ \mu_1 \sin i_c = \frac{\mu_1}{2} \]
- Dividing both sides by \( \mu_1 \) (assuming \( \mu_1 \neq 0 \)):
\[ \sin i_c = \frac{1}{2} \]

5. **Finding the Critical Angle**:
- The angle whose sine is \( \frac{1}{2} \) is \( 30^\circ \):
\[ i_c = 30^\circ \]

6. **Conclusion**:
- For total internal reflection to occur, the angle of incidence \( i \) must be greater than the critical angle \( i_c \):
\[ i > 30^\circ \]

### Final Answer:
The angle of incidence must be greater than \( 30^\circ \).

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