A luminous point object is moving along the principal axis of a concave mirror of focal length 12 cm towards it. When its distance from the mirror is 20 cm its velocity is 4 cm/s. the velocity of the image in cm/s at that instant is

Options

A. 6, towards the mirror
B. 6, away from the mirror
C. 9, away from the mirror
D. 9,towards the mirror

Answer

To solve the problem step by step, we will use the mirror formula and the concept of image velocity in relation to object velocity.

### Step 1: Understand the given values
- Focal length of the concave mirror (f) = -12 cm (negative because it is a concave mirror).
- Object distance (u) = -20 cm (negative as per the sign convention).
- Object velocity (u') = +4 cm/s (positive since the object is moving towards the mirror).

### Step 2: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{-12} = \frac{1}{v} + \frac{1}{-20} \]

### Step 3: Solve for v (image distance)
Rearranging the equation:
\[ \frac{1}{v} = \frac{1}{-12} + \frac{1}{20} \]
Finding a common denominator (which is 60):
\[ \frac{1}{v} = \frac{-5}{60} + \frac{3}{60} = \frac{-2}{60} \]
Thus,
\[ v = \frac{60}{-2} = -30 \text{ cm} \]
This means the image is formed 30 cm on the same side as the object.

### Step 4: Differentiate the mirror formula
To find the velocity of the image (v'), we differentiate the mirror formula with respect to time:
\[ \frac{d}{dt}\left(\frac{1}{f}\right) = \frac{d}{dt}\left(\frac{1}{v}\right) + \frac{d}{dt}\left(\frac{1}{u}\right) \]
Since f is constant, its derivative is zero:
\[ 0 = -\frac{1}{v^2} \cdot v' - \frac{1}{u^2} \cdot u' \]

### Step 5: Substitute known values
Substituting the values we have:
- \( v = -30 \) cm
- \( u = -20 \) cm
- \( u' = 4 \) cm/s

Now substituting into the differentiated equation:
\[ 0 = -\frac{1}{(-30)^2} \cdot v' - \frac{1}{(-20)^2} \cdot 4 \]
This simplifies to:
\[ 0 = -\frac{1}{900} v' + \frac{4}{400} \]
\[ 0 = -\frac{1}{900} v' + \frac{1}{100} \]

### Step 6: Solve for v'
Rearranging gives:
\[ \frac{1}{900} v' = \frac{1}{100} \]
Multiplying both sides by 900:
\[ v' = 900 \cdot \frac{1}{100} = 9 \text{ cm/s} \]

### Step 7: Determine the direction of the image velocity
Since the image is moving away from the mirror, the velocity of the image is +9 cm/s.

### Final Answer:
The velocity of the image at that instant is **9 cm/s away from the mirror**.

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