A spherical mirror from image at distance 10 cm of an object at distance 30 cm from mirror. If object start moving with velocity 9cm/sec. Find velocity of image

Options

A. -9cm
B. -4cm
C. -1cm
D. -3cm

Answer

To solve the problem, we will use the mirror formula and the concept of differentiation to find the velocity of the image when the object is moving.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Object distance (u) = -30 cm (the negative sign is used because the object is in front of the mirror)
- Image distance (v) = 10 cm (positive because the image is formed on the same side as the object)
- Object velocity (du/dt) = 9 cm/s (positive since the object is moving away from 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. Plugging in the values:
\[ \frac{1}{f} = \frac{1}{10} + \frac{1}{-30} \]
To solve this, find a common denominator:
\[ \frac{1}{f} = \frac{3}{30} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15} \]
Thus, the focal length \( f = 15 \) cm.

3. **Differentiate the Mirror Formula:**
Differentiate the mirror formula with respect to time (t):
\[ \frac{d}{dt}\left(\frac{1}{v}\right) + \frac{d}{dt}\left(\frac{1}{u}\right) = 0 \]
Using the chain rule:
\[ -\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0 \]

4. **Rearranging the Equation:**
Rearranging gives:
\[ \frac{dv}{dt} = -\frac{v^2}{u^2} \cdot \frac{du}{dt} \]

5. **Substituting the Known Values:**
Substitute \( v = 10 \) cm, \( u = -30 \) cm, and \( \frac{du}{dt} = 9 \) cm/s into the equation:
\[ \frac{dv}{dt} = -\frac{(10)^2}{(-30)^2} \cdot 9 \]
Simplifying:
\[ \frac{dv}{dt} = -\frac{100}{900} \cdot 9 = -\frac{1}{9} \cdot 9 = -1 \text{ cm/s} \]

6. **Conclusion:**
The velocity of the image is \( \frac{dv}{dt} = -1 \) cm/s. The negative sign indicates that the image is moving in the opposite direction to the object.

### Final Answer:
The velocity of the image is \( -1 \) cm/s.

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