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 `9m s^(-1)` the speed with which the image moves is

Options

A. `10 ms^(-1)`
B. `1 ms^(-1)`
C. `9 ms^(-1)`
D. `0.9 ms^(-1)`

Explanation

To solve the problem step by step, we will use the mirror formula and differentiate it to find the relationship between the speeds of the object and the image.

### Step 1: Understand the given information
- The object distance (u) is given as 30 cm (which we will take as -30 cm for concave mirrors, as per the sign convention).
- The image distance (v) is given as 10 cm (which we will take as -10 cm for real images formed by concave mirrors).
- The speed of the object (du/dt) is given as 9 m/s.

### Step 2: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
However, for this problem, we will differentiate the mirror equation instead.

### Step 3: Differentiate the mirror equation
Differentiating the equation with respect to time (t):
\[ -\frac{1}{u^2} \frac{du}{dt} - \frac{1}{v^2} \frac{dv}{dt} = 0 \]
Rearranging gives us:
\[ \frac{dv}{dt} = -\frac{v^2}{u^2} \frac{du}{dt} \]

### Step 4: Substitute the known values
We know:
- \( u = -30 \, \text{cm} \)
- \( v = -10 \, \text{cm} \)
- \( \frac{du}{dt} = 9 \, \text{m/s} = 900 \, \text{cm/s} \) (since we need to keep the units consistent)

Now substituting these values into the differentiated equation:
\[ \frac{dv}{dt} = -\frac{(-10)^2}{(-30)^2} \cdot 900 \]
Calculating:
\[ = -\frac{100}{900} \cdot 900 \]
\[ = -100 \, \text{cm/s} \]

### Step 5: Convert to meters per second
Since the question asks for the speed in meters per second:
\[ \frac{dv}{dt} = -1 \, \text{m/s} \]

### Conclusion
The speed with which the image moves is \( 1 \, \text{m/s} \) in the opposite direction of the object.

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