A ray of light falls on a plane mirror. When the mirror is turned, about an axis at right angles to the plane of mirror by `20^(@)` the angle between the incident ray and new reflected ray is `45^(@)`. The angle between the incident ray and original reflected ray was therefore.

Options

A. (a) `35^ (@)` or `50^ (@)`
B. (b) `25^(@)` or `65^(@)`
C. (c) `45^(@)` or `5^(@)`
D. (d) None of these

Explanation

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the initial setup
When a ray of light strikes a plane mirror, it does so at a certain angle of incidence (I). According to the law of reflection, the angle of reflection (R) is equal to the angle of incidence. Thus, we have:
\[ I = R \]

### Step 2: Consider the rotation of the mirror
The problem states that the mirror is turned by \(20^\circ\) about an axis that is perpendicular to its plane. This means that the normal to the mirror surface also rotates by \(20^\circ\).

### Step 3: Analyze the effect of the mirror's rotation
When the mirror is rotated by \(20^\circ\), the new angle of incidence (let's call it \(I'\)) will be the original angle of incidence (I) plus the angle of rotation:
\[ I' = I + 20^\circ \]

The angle of reflection will also change accordingly:
\[ R' = I' = I + 20^\circ \]

### Step 4: Determine the relationship between the new reflected ray and the original reflected ray
According to the problem, the angle between the incident ray and the new reflected ray is \(45^\circ\). This means:
\[ I' + R' = 45^\circ \]

Since \(R' = I + 20^\circ\), we can substitute this into the equation:
\[ I + 20^\circ + I + 20^\circ = 45^\circ \]
\[ 2I + 40^\circ = 45^\circ \]

### Step 5: Solve for the original angle of incidence
Now, we can solve for \(I\):
\[ 2I = 45^\circ - 40^\circ \]
\[ 2I = 5^\circ \]
\[ I = \frac{5^\circ}{2} = 2.5^\circ \]

### Step 6: Find the angle between the incident ray and the original reflected ray
Since the angle of reflection \(R\) is equal to the angle of incidence \(I\):
\[ R = I = 2.5^\circ \]

The angle between the incident ray and the original reflected ray is:
\[ I + R = 2.5^\circ + 2.5^\circ = 5^\circ \]

### Final Answer
The angle between the incident ray and the original reflected ray is \(5^\circ\).

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