An object placed in front of a plane mirror is displaced by 0.4 m along a straight line at an angle of `30^(@)` to mirror plane. The change in the distance between the object and its image is:

Options

A. 0.20 m
B. 0.40 m
C. 0.25 m
D. 0.80 m

Answer

To solve the problem of finding the change in distance between an object and its image after the object is displaced, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Setup**:
- Let the initial distance of the object from the mirror be \( u \).
- The distance between the object and its image in a plane mirror is given by \( d = d_i - d_o = 2u \).

2. **Displacement of the Object**:
- The object is displaced by \( 0.4 \, \text{m} \) at an angle of \( 30^\circ \) to the mirror plane.

3. **Calculate the Horizontal and Vertical Components of Displacement**:
- The horizontal component of the displacement (along the mirror plane) can be calculated using:
\[ \Delta x = 0.4 \cos(30^\circ) \]
- The vertical component is not relevant for the distance between the object and its image.

4. **Calculate the Horizontal Component**:
- Using \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ \Delta x = 0.4 \times \frac{\sqrt{3}}{2} = 0.4 \times 0.866 = 0.3464 \, \text{m} \]

5. **New Distance Between Object and Image**:
- The new distance between the object and its image after displacement becomes:
\[ d' = 2(u + \Delta x) = 2u + 2 \Delta x \]

6. **Change in Distance**:
- The change in distance \( \Delta d \) is given by:
\[ \Delta d = d' - d = (2u + 2\Delta x) - 2u = 2\Delta x \]
- Substituting \( \Delta x \):
\[ \Delta d = 2 \times 0.3464 \approx 0.6928 \, \text{m} \]

7. **Final Result**:
- The total change in distance between the object and its image is approximately \( 0.6928 \, \text{m} \).