A beam of light from a source L is incident normally on a plane mirror fixed at a certin distance x from the source. The beam is reflacted back as a spot on a scale placed just above the surce L. When the mirror is rotated through a small angle `theta` the spot of the light is found to move through a distance y on the scale. The angle `theta` is given by

Options

A. `y/(2x)`
B. `y/(x)`
C. `x/(2y)`
D. `x/(y)`

Answer

To solve the problem, we need to analyze the situation involving the plane mirror and the light beam. Here’s a step-by-step solution:

### Step 1: Understand the Setup
A beam of light from a source \( L \) is incident normally on a plane mirror. The mirror is at a distance \( x \) from the source. When the mirror is rotated through a small angle \( \theta \), the reflected light spot moves a distance \( y \) on a scale above the source.

### Step 2: Analyze the Reflection
When the mirror is rotated through an angle \( \theta \), the angle of reflection changes. According to the law of reflection, the angle of incidence equals the angle of reflection. Therefore, if the mirror is rotated by an angle \( \theta \), the light beam will be deflected by an angle \( 2\theta \) (since it reflects off the mirror).

### Step 3: Relate the Movement of the Spot to the Angle
The distance \( y \) that the spot moves on the scale can be related to the angle \( 2\theta \) using trigonometric relationships. For small angles, we can use the tangent function:
\[ \tan(2\theta) \approx 2\theta \quad \text{(for small angles)} \]
This gives us:
\[ \tan(2\theta) = \frac{y}{x} \]

### Step 4: Substitute and Simplify
Using the approximation for small angles:
\[ 2\theta \approx \frac{y}{x} \]
Now, we can solve for \( \theta \):
\[ \theta \approx \frac{y}{2x} \]

### Step 5: Conclusion
Thus, the angle \( \theta \) is given by:
\[ \theta = \frac{y}{2x} \]

### Final Answer
The angle \( \theta \) is \( \frac{y}{2x} \).

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