When an object is kept at a distance of 30cm from a concave mirror, the image is formed at a distance of 10 cm. If the object is moved with a speed of `9cm s^(-1)` the speed with which the image moves is

Options

A. `0.1ms^(-1)`
B. `1ms^(-1)`
C. `3ms^(-1)`
D. `9ms^(-1)`

Explanation

To solve the problem, we will use the mirror formula and the concept of differentiation with respect to time. Here are the steps to find the speed of the image when the object is moved:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Object distance (U) = -30 cm (the negative sign indicates that the object is in front of the mirror)
- Image distance (V) = -10 cm (the negative sign indicates that the image is real and formed in front of the mirror)
- Speed of the object (dU/dt) = -9 cm/s (the negative sign indicates that the object is moving towards the mirror)

2. **Use the Mirror Formula:**
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{V} + \frac{1}{U} \]
We need to find the focal length (f) first.

Substituting the values:
\[ \frac{1}{f} = \frac{1}{-10} + \frac{1}{-30} \]
Finding a common denominator (which is 30):
\[ \frac{1}{f} = -\frac{3}{30} - \frac{1}{30} = -\frac{4}{30} = -\frac{2}{15} \]
Thus,
\[ f = -\frac{15}{2} = -7.5 \text{ cm} \]

3. **Differentiate the Mirror Formula with Respect to Time (t):**
Differentiating both sides of the mirror formula with respect to time gives:
\[ 0 = -\frac{dV}{V^2} \frac{dV}{dt} - \frac{dU}{U^2} \frac{dU}{dt} \]
Rearranging gives:
\[ \frac{dV}{dt} = -\frac{V^2}{U^2} \frac{dU}{dt} \]

4. **Substitute the Known Values:**
Now substitute U = -30 cm, V = -10 cm, and dU/dt = -9 cm/s into the equation:
\[ \frac{dV}{dt} = -\frac{(-10)^2}{(-30)^2} \cdot (-9) \]
Simplifying this:
\[ \frac{dV}{dt} = -\frac{100}{900} \cdot (-9) = \frac{1}{9} \cdot 9 = 1 \text{ cm/s} \]

5. **Conclusion:**
The speed with which the image moves is:
\[ \frac{dV}{dt} = 1 \text{ cm/s} \]

### Final Answer:
The speed with which the image moves is **1 cm/s**.