A particle moves towards a concave mirror of focal length 30 cm along its axis and with a constant speed of 4 cm/ sec. What is the speed of its image when the particle is at 90 cm from the mirror?

Options

A. 2 cm/s
B. 8 cm/s
C. 1 cm/s
D. 4 cm/s

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given values
- Focal length of the concave mirror, \( f = -30 \, \text{cm} \) (negative because it is a concave mirror).
- Object distance, \( u = -90 \, \text{cm} \) (negative as per sign convention for mirrors).
- Speed of the object, \( V_o = 4 \, \text{cm/s} \).

### Step 2: Use the mirror formula to find the image distance \( v \)
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
Substituting the known values:
\[ \frac{1}{-30} = \frac{1}{-90} + \frac{1}{v} \]

### Step 3: Solve for \( \frac{1}{v} \)
Rearranging the equation:
\[ \frac{1}{v} = \frac{1}{-30} - \frac{1}{-90} \]
Finding a common denominator (which is 90):
\[ \frac{1}{v} = \frac{-3 + 1}{90} = \frac{-2}{90} \]
Thus, we have:
\[ v = -45 \, \text{cm} \]

### Step 4: Differentiate the mirror formula with respect to time
Differentiating the mirror formula:
\[ \frac{d}{dt}\left(\frac{1}{f}\right) = \frac{d}{dt}\left(\frac{1}{u}\right) + \frac{d}{dt}\left(\frac{1}{v}\right) \]
Since \( f \) is constant, \( \frac{d}{dt}\left(\frac{1}{f}\right) = 0 \):
\[ 0 = -\frac{1}{u^2} \frac{du}{dt} - \frac{1}{v^2} \frac{dv}{dt} \]

### Step 5: Substitute known values
Let \( V_i \) be the speed of the image:
\[ 0 = -\frac{1}{u^2} V_o - \frac{1}{v^2} V_i \]
Rearranging gives:
\[ \frac{1}{v^2} V_i = \frac{1}{u^2} V_o \]
Thus:
\[ V_i = \frac{v^2}{u^2} V_o \]

### Step 6: Substitute \( u \), \( v \), and \( V_o \)
Substituting the values:
\[ V_i = \frac{(-45)^2}{(-90)^2} \times 4 \]
Calculating:
\[ V_i = \frac{2025}{8100} \times 4 = \frac{1}{4} \times 4 = 1 \, \text{cm/s} \]

### Step 7: Conclusion
The speed of the image is \( 1 \, \text{cm/s} \).

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