A ray of light is incident at an angle `alpha` on the boundary separating two transparant media. It transmited in other medium. If the angle incidence is increased very slightly, the ray gets reflected in the same medium. The different between angles of deviation in the two cases will close to

Options

A. `2alpha`
B. `90^(@)-alpha`
C. `180^(@)-alpha`
D. `180^(@)-alpha`

Answer

To solve the problem, we need to analyze the situation of light rays incident at an angle \(\alpha\) on the boundary of two transparent media and how the deviation changes when the angle of incidence is slightly increased.

### Step-by-Step Solution:

1. **Understanding the Critical Angle**:
- When a ray of light is incident at an angle \(\alpha\) and it transmits into another medium, we can infer that \(\alpha\) is the critical angle. This means that the angle of refraction is \(90^\circ\) in the second medium.

2. **Calculating Deviation for the First Case**:
- When the light ray is incident at the critical angle \(\alpha\), the angle of deviation (\(\delta_1\)) can be calculated as:
\[ \delta_1 = 90^\circ - \alpha \]
- This is because the ray refracts at \(90^\circ\) to the normal.

3. **Increasing the Angle of Incidence**:
- If we increase the angle of incidence slightly beyond the critical angle, the ray will no longer transmit into the second medium but will instead reflect back into the first medium.

4. **Calculating Deviation for the Second Case**:
- In this case, the angle of incidence becomes slightly greater than \(\alpha\). Let's denote this new angle as \(\alpha + \epsilon\) (where \(\epsilon\) is a very small angle).
- The angle of reflection will also be \(\alpha + \epsilon\).
- The angle of deviation (\(\delta_2\)) for this case can be calculated as:
\[ \delta_2 = 180^\circ - 2(\alpha + \epsilon) \]
- Simplifying this gives:
\[ \delta_2 = 180^\circ - 2\alpha - 2\epsilon \]

5. **Finding the Difference in Angles of Deviation**:
- Now, we can find the difference in the angles of deviation between the two cases:
\[ \Delta \delta = \delta_2 - \delta_1 \]
- Substituting the values we calculated:
\[ \Delta \delta = (180^\circ - 2\alpha - 2\epsilon) - (90^\circ - \alpha) \]
- Simplifying this expression:
\[ \Delta \delta = 180^\circ - 2\alpha - 2\epsilon - 90^\circ + \alpha \]
\[ \Delta \delta = 90^\circ - \alpha - 2\epsilon \]

6. **Conclusion**:
- As \(\epsilon\) is very small, we can approximate:
\[ \Delta \delta \approx 90^\circ - \alpha \]
- Thus, the difference between the angles of deviation in the two cases is approximately \(90^\circ - \alpha\).

### Final Answer:
The difference between the angles of deviation in the two cases will be close to \(90^\circ - \alpha\).