A point object is moving on the principal axis of a concave mirror of focal length `24 cm` towards the mirror. When it is at a distance of `60 cm` from the mirror, its velocity is . `9 sec// cm` What is the velocity of the image at that instant

Options

A. 5 cm/sec towards the mirror
B. 4 cm/sec towards the mirror
C. 4 cm/sec away from the mirror
D. 9 cm/sec away from the mirror

Answer

To solve the problem step by step, we will use the mirror formula and the concept of related rates in calculus.

### Step 1: Identify the given values
- Focal length of the concave mirror, \( f = -24 \, \text{cm} \) (negative because it is a concave mirror)
- Object distance, \( u = -60 \, \text{cm} \) (negative as per the sign convention for mirrors)
- Velocity of the object, \( \frac{du}{dt} = -9 \, \text{cm/s} \) (negative because the object is moving towards the mirror)

### Step 2: Use the mirror formula to find the image distance \( v \)
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{-24} = \frac{1}{v} + \frac{1}{-60} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{-24} + \frac{1}{60} \]
Finding a common denominator (which is 120):
\[ \frac{1}{v} = \frac{-5}{120} + \frac{2}{120} = \frac{-3}{120} \]
Thus,
\[ v = -40 \, \text{cm} \]
(The negative sign indicates that the image is formed on the same side as the object, which is typical for a concave mirror.)

### Step 3: Differentiate the mirror formula
To find the velocity of the image, we differentiate the mirror formula with respect to time \( t \):
\[ \frac{d}{dt}\left(\frac{1}{f}\right) = \frac{d}{dt}\left(\frac{1}{v}\right) + \frac{d}{dt}\left(\frac{1}{u}\right) \]
Since \( f \) is constant, \( \frac{d}{dt}\left(\frac{1}{f}\right) = 0 \). Thus:
\[ 0 = -\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} \]
Rearranging gives:
\[ \frac{dv}{dt} = -\frac{v^2}{u^2} \frac{du}{dt} \]

### Step 4: Substitute known values
We already found \( v = -40 \, \text{cm} \) and \( u = -60 \, \text{cm} \). Now substituting these values along with \( \frac{du}{dt} = -9 \, \text{cm/s} \):
\[ \frac{dv}{dt} = -\frac{(-40)^2}{(-60)^2} \cdot (-9) \]
Calculating the squares:
\[ = -\frac{1600}{3600} \cdot (-9) \]
Simplifying:
\[ = \frac{1600 \cdot 9}{3600} = \frac{14400}{3600} = 4 \, \text{cm/s} \]

### Step 5: Determine the direction of the image velocity
Since the image velocity \( \frac{dv}{dt} = 4 \, \text{cm/s} \) is positive, it indicates that the image is moving away from the mirror.

### Final Answer
The velocity of the image at that instant is \( 4 \, \text{cm/s} \) away from the mirror.

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