A point object is placed at a distance of 10 cm and its real image is formed at a distance of 20 cm from a concave mirror. If the object is moved by 0.1 cm towards the mirror, the image will shift by about

Options

A. 0.4 cm away from the mirror
B. 0.8 cm away from the mirror
C. 0.4 cm towards the mirror
D. 0.8 cm towards the mirror

Explanation

To solve the problem, we will use the mirror formula and the concept of differentiation to find how the image distance changes when the object distance is altered.

### Step-by-step Solution:

1. **Identify Given Values:**
- Object distance (u) = -10 cm (the negative sign indicates that the object is in front of the mirror).
- Image distance (v) = 20 cm (positive since it is a real image).
- Change in object distance (du) = -0.1 cm (the object is moved towards the mirror).

2. **Use the Mirror Formula:**
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
where \( f \) is the focal length of the mirror.

3. **Differentiate the Mirror Formula:**
Differentiating both sides with respect to time (or any variable), we get:
\[ 0 = \frac{d}{dt}\left(\frac{1}{v}\right) + \frac{d}{dt}\left(\frac{1}{u}\right) \]
This can be expressed as:
\[ 0 = -\frac{1}{v^2} \cdot dv - \frac{1}{u^2} \cdot du \]
Rearranging gives us:
\[ \frac{1}{v^2} \cdot dv = -\frac{1}{u^2} \cdot du \]

4. **Solve for dv:**
From the rearranged equation, we can express \( dv \):
\[ dv = -\frac{v^2}{u^2} \cdot du \]

5. **Substitute the Known Values:**
Substitute \( v = 20 \) cm, \( u = -10 \) cm, and \( du = -0.1 \) cm into the equation:
\[ dv = -\frac{(20)^2}{(-10)^2} \cdot (-0.1) \]
Simplifying this gives:
\[ dv = -\frac{400}{100} \cdot (-0.1) = 4 \cdot 0.1 = 0.4 \text{ cm} \]

6. **Interpret the Result:**
Since \( dv \) is positive, it indicates that the image moves away from the mirror. Therefore, the image shifts by approximately 0.4 cm away from the mirror.

### Final Answer:
The image will shift by about **0.4 cm** away from the mirror.