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^(@)`

Explanation

To solve the problem of determining the angle of incidence for total internal reflection when light transitions from medium 1 to medium 2, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two media: medium 1 and medium 2.
- The velocity of light in medium 2 (v2) is double that in medium 1 (v1), i.e., \( v_2 = 2v_1 \).

2. **Finding the Refractive Indices**:
- The refractive index (μ) of a medium is defined as the ratio of the speed of light in vacuum (c) to the speed of light in that medium.
- For medium 1, the refractive index \( \mu_1 = \frac{c}{v_1} \).
- For medium 2, the refractive index \( \mu_2 = \frac{c}{v_2} = \frac{c}{2v_1} = \frac{1}{2} \cdot \frac{c}{v_1} = \frac{1}{2} \mu_1 \).

3. **Calculating the Critical Angle**:
- Total internal reflection occurs when the angle of incidence exceeds the critical angle (ic).
- The critical angle can be found using Snell's law:
\[ \mu_1 \sin(i_c) = \mu_2 \sin(90^\circ) \]
- Since \( \sin(90^\circ) = 1 \), we can rewrite this as:
\[ \mu_1 \sin(i_c) = \mu_2 \]

4. **Substituting the Refractive Indices**:
- We know \( \mu_2 = \frac{1}{2} \mu_1 \), so substituting this into the equation gives:
\[ \mu_1 \sin(i_c) = \frac{1}{2} \mu_1 \]
- Dividing both sides by \( \mu_1 \) (assuming \( \mu_1 \neq 0 \)):
\[ \sin(i_c) = \frac{1}{2} \]

5. **Finding the Critical Angle**:
- Now we can find the critical angle \( i_c \):
\[ i_c = \sin^{-1}\left(\frac{1}{2}\right) \]
- This gives us:
\[ i_c = 30^\circ \]

6. **Conclusion**:
- For total internal reflection to occur, the angle of incidence must be greater than the critical angle:
\[ \text{Angle of incidence must be greater than } 30^\circ. \]

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

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