Light falls from glass `(mu=1.5)` to air. Find the angle of incidence for which the angle of deviation is `90^@`.

Explanation

To solve the problem of finding the angle of incidence for which the angle of deviation is 90 degrees when light passes from glass (with a refractive index, µ = 1.5) to air, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Refractive Indices**:
- The refractive index of glass (µ_glass) = 1.5
- The refractive index of air (µ_air) = 1 (approximately)

2. **Calculate the Critical Angle (θ_C)**:
- The critical angle can be calculated using Snell's law. The formula for the critical angle is:
\[ \sin(θ_C) = \frac{µ_{air}}{µ_{glass}} = \frac{1}{1.5} = \frac{2}{3} \]
- Therefore,
\[ θ_C = \sin^{-1}\left(\frac{2}{3}\right) \]
- Using a calculator, we find:
\[ θ_C \approx 41.8^\circ \]

3. **Determine the Maximum Deviation**:
- The maximum deviation occurs when the angle of incidence is equal to the critical angle. The maximum angle of deviation (D_max) in this case can be calculated as:
\[ D_{max} = 90^\circ - θ_C = 90^\circ - 41.8^\circ = 48.2^\circ \]
- However, since we are looking for a deviation of 90 degrees, we note that total internal reflection must occur.

4. **Relate Deviation to Angle of Incidence**:
- The angle of deviation (D) for refraction is given by:
\[ D = 180^\circ - 2I \]
- For our case, we set D = 90 degrees:
\[ 90^\circ = 180^\circ - 2I \]
- Rearranging gives:
\[ 2I = 180^\circ - 90^\circ = 90^\circ \]
- Thus,
\[ I = \frac{90^\circ}{2} = 45^\circ \]

5. **Conclusion**:
- The angle of incidence (I) for which the angle of deviation is 90 degrees is:
\[ I = 45^\circ \]

### Final Answer:
The angle of incidence for which the angle of deviation is 90 degrees is **45 degrees**.