A particle moves toward a concave mirror of focal length 30 cm along its axis and with a constant speed of `(4 cm)/(s)`. 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 procedures:

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

### Step 2: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]
Substituting the known values into the formula:
\[ \frac{1}{v} + \frac{1}{-90} = \frac{1}{-30} \]
This simplifies to:
\[ \frac{1}{v} - \frac{1}{90} = -\frac{1}{30} \]

### Step 3: Solve for \( v \)
Rearranging the equation:
\[ \frac{1}{v} = -\frac{1}{30} + \frac{1}{90} \]
Finding a common denominator (which is 90):
\[ \frac{1}{v} = -\frac{3}{90} + \frac{1}{90} = -\frac{2}{90} = -\frac{1}{45} \]
Thus:
\[ v = -45 \, \text{cm} \]
This means the image is formed at a distance of 45 cm on the same side as the object (real image).

### Step 4: Differentiate the mirror formula
We 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:
\[ -\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0 \]
Rearranging gives:
\[ \frac{dv}{dt} = -\frac{v^2}{u^2} \frac{du}{dt} \]

### Step 5: Substitute the known values
Substituting \( v = -45 \, \text{cm} \), \( u = -90 \, \text{cm} \), and \( \frac{du}{dt} = 4 \, \text{cm/s} \):
\[ \frac{dv}{dt} = -\frac{(-45)^2}{(-90)^2} \cdot 4 \]
Calculating:
\[ \frac{dv}{dt} = -\frac{2025}{8100} \cdot 4 = -\frac{1}{4} \cdot 4 = -1 \, \text{cm/s} \]

### Step 6: Find the speed of the image
The speed of the image is given by the magnitude of \( \frac{dv}{dt} \):
\[ \text{Speed of the image} = 1 \, \text{cm/s} \]

### Final Answer
The speed of the image when the particle is at 90 cm from the mirror is **1 cm/s**.

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