A point object on the principal axis at a distance 1.5 cm in front of concave mirror of radius of curvature 20 cm has velocity 2mm/s is perpendicular to the principal axis. The velocity of image at that instant will be

Options

A. 2.34mm/s
B. 4 mm/s
C. 8.26 mm/s
D. 16 mm/s

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Distance of the object from the mirror (u) = -1.5 cm (negative because the object is in front of the mirror)
- Radius of curvature (R) = 20 cm
- Focal length (f) = R/2 = -10 cm (negative for concave mirror)
- Velocity of the object (v_o) = 2 mm/s (perpendicular to the principal axis)

### Step 2: Use the mirror formula to find the image distance (v)
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Rearranging this gives:
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{v} = \frac{1}{-10} - \frac{1}{-1.5} \]
Calculating the right-hand side:
\[ \frac{1}{v} = -0.1 + 0.6667 = 0.5667 \]
Thus,
\[ v = \frac{1}{0.5667} \approx 1.76 \text{ cm} \]

### Step 3: Convert image distance to mm
Since the answer needs to be in mm:
\[ v \approx 17.6 \text{ mm} \]

### Step 4: Relate the heights of the object and image using magnification
The magnification (m) is given by:
\[ m = -\frac{v}{u} = \frac{h_i}{h_o} \]
Where \(h_i\) is the height of the image and \(h_o\) is the height of the object.

### Step 5: Differentiate the magnification equation with respect to time
Differentiating gives:
\[ \frac{dh_i}{dt} = -\frac{v}{u} \frac{dh_o}{dt} \]
Let \(v_i\) be the velocity of the image and \(v_o\) be the velocity of the object:
\[ v_i = -\frac{v}{u} v_o \]

### Step 6: Substitute the known values
Substituting \(v = 17.6 \text{ mm}\), \(u = 15 \text{ mm}\), and \(v_o = 2 \text{ mm/s}\):
\[ v_i = -\frac{17.6}{15} \times 2 \]
Calculating:
\[ v_i = -1.1733 \times 2 \approx -2.34 \text{ mm/s} \]

### Step 7: Interpret the result
The negative sign indicates that the image is moving in the opposite direction to the object, which is downward.

### Final Answer
The velocity of the image is approximately **2.34 mm/s downward**.

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