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 use the principles of reflection and refraction of light.

### Step-by-Step Solution:

1. **Understand the Problem Statement**:
We have a ray of light falling on a glass plate with a refractive index (μ) of 1.5. We need to find the angle of incidence (I) when the angle between the reflected ray and the refracted ray is 90 degrees.

2. **Identify the Angles**:
- Let the angle of incidence be \( I \).
- According to the law of reflection, the angle of reflection \( R \) is equal to the angle of incidence, so \( R = I \).
- The angle of refraction is denoted as \( r \).

3. **Set Up the Relationship Between Angles**:
Since the angle between the reflected ray and the refracted ray is 90 degrees, we can express this relationship as:
\[ R + r = 90^\circ \]
Substituting \( R \) with \( I \):
\[ I + r = 90^\circ \]
Therefore, we can express \( r \) as:
\[ r = 90^\circ - I \]

4. **Apply Snell's Law**:
Snell's Law states:
\[ \mu_1 \sin I = \mu_2 \sin r \]
Where:
- \( \mu_1 \) (refractive index of air) = 1
- \( \mu_2 \) (refractive index of glass) = 1.5

Substituting the values we have:
\[ 1 \cdot \sin I = 1.5 \cdot \sin(90^\circ - I) \]

5. **Use the Co-function Identity**:
Using the identity \( \sin(90^\circ - I) = \cos I \), we can rewrite the equation:
\[ \sin I = 1.5 \cdot \cos I \]

6. **Rearranging the Equation**:
Dividing both sides by \( \cos I \):
\[ \tan I = 1.5 \]

7. **Calculate the Angle of Incidence**:
To find \( I \), take the arctangent:
\[ I = \tan^{-1}(1.5) \]
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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