A point object P moves towards a convex mirror with a constant speed v, along its optic axis.The speed of the image

Options

A. Is always less than v
B. May be greater than and equal to v or less than v depending on the position of P
C. Increases as P comes closer to the mirror
D. Decreases as P comes closer to the mirror

Answer

To find the speed of the image formed by a convex mirror when a point object P moves towards it with a constant speed \( v \), we can follow these steps:

### Step 1: Understand the Mirror Formula
The mirror formula for a convex mirror is given by:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]
where:
- \( v \) = image distance (positive for virtual images in convex mirrors)
- \( u \) = object distance (negative for real objects in front of the mirror)
- \( f \) = focal length (positive for convex mirrors)

### Step 2: Differentiate the Mirror Formula
To find the speed of the image, we need to differentiate the mirror formula with respect to time \( t \):
\[ \frac{d}{dt}\left(\frac{1}{v}\right) + \frac{d}{dt}\left(\frac{1}{u}\right) = 0 \]
This gives us:
\[ -\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0 \]

### Step 3: Rearranging the Equation
Rearranging the differentiated equation leads to:
\[ \frac{dv}{dt} = -\frac{u^2}{v^2} \frac{du}{dt} \]
Here, \( \frac{du}{dt} \) is the speed of the object, which is given as \( v_0 \) (the constant speed of the object moving towards the mirror).

### Step 4: Substitute the Values
Substituting \( \frac{du}{dt} = -v_0 \) (negative because \( u \) is decreasing as the object approaches the mirror):
\[ \frac{dv}{dt} = \frac{u^2}{v^2} v_0 \]
This indicates that the speed of the image \( \frac{dv}{dt} \) is positive, meaning the image is moving away from the mirror.

### Step 5: Analyze Image Speed
As the object approaches the mirror, \( u \) decreases. The image speed \( \frac{dv}{dt} \) will increase as \( u \) decreases, indicating that the image moves faster as the object gets closer to the mirror.

### Step 6: Conclusion
The speed of the image is always less than the speed of the object. As the object approaches the mirror, the speed of the image increases, reaching a maximum as the object approaches the focal point of the mirror.

### Final Result
The speed of the image \( \frac{dv}{dt} \) can be expressed as:
\[ \frac{dv}{dt} = \frac{u^2}{v^2} v_0 \]
where \( v_0 \) is the speed of the object.

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