A ray of light falls on a glass plate of refractive index `mu=1.5`.
What is the angle of incidence of the ray if the angle between the reflected and
refracted rays is `90^@`?

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a ray of light incident on a glass plate with a refractive index \( \mu = 1.5 \).
- The angle between the reflected ray and the refracted ray is given as \( 90^\circ \).

2. **Define Angles**:
- Let the angle of incidence be \( I \).
- According to the law of reflection, the angle of reflection is also \( I \).
- Let the angle of refraction be \( R \).

3. **Using the Given Information**:
- The angle between the reflected ray and the refracted ray is \( 90^\circ \).
- Therefore, we can write the relationship:
\[ R + I = 90^\circ \]
- From this, we can express \( R \) in terms of \( I \):
\[ R = 90^\circ - I \]

4. **Applying Snell's Law**:
- Snell's law states:
\[ \mu_1 \sin I = \mu_2 \sin R \]
- Here, \( \mu_1 = 1 \) (air) and \( \mu_2 = 1.5 \) (glass).
- Substituting the values into Snell's law:
\[ 1 \cdot \sin I = 1.5 \cdot \sin R \]

5. **Substituting for \( R \)**:
- We substitute \( R = 90^\circ - I \) into the equation:
\[ \sin I = 1.5 \cdot \sin(90^\circ - I) \]
- Using the identity \( \sin(90^\circ - \theta) = \cos \theta \):
\[ \sin I = 1.5 \cdot \cos I \]

6. **Rearranging the Equation**:
- We can rearrange this to:
\[ \frac{\sin I}{\cos I} = 1.5 \]
- This simplifies to:
\[ \tan I = 1.5 \]

7. **Finding the Angle of Incidence**:
- To find \( I \), we take the inverse tangent:
\[ I = \tan^{-1}(1.5) \]
- Using a calculator, we find:
\[ I \approx 56.3^\circ \]

### Final Answer:
The angle of incidence \( I \) is approximately \( 56.3^\circ \).

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