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 `lt` V
B. may be `gt`, = or `lt` V depending on the position of P. .
C. increase as P comes closer to the mirror
D. decrease as P comes closer to the mirror

Answer

To solve the problem of determining the speed of the image formed by a convex mirror when a point object 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:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
where:
- \( f \) is the focal length of the mirror (negative for convex mirrors),
- \( v \) is the image distance (positive for virtual images),
- \( u \) is the object distance (negative as the object is in front of the mirror).

### Step 2: Set Up the Variables
Let:
- \( x \) be the distance of the object from the mirror (so \( u = -x \)),
- \( y \) be the distance of the image from the mirror.

### Step 3: Differentiate the Mirror Formula
Differentiating 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, its derivative is zero:
\[ 0 = \frac{d}{dt}\left(\frac{1}{v}\right) + \frac{d}{dt}\left(\frac{1}{-x}\right) \]

### Step 4: Apply the Chain Rule
Using the chain rule, we have:
\[ 0 = -\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{x^2} \frac{dx}{dt} \]
Rearranging gives:
\[ \frac{dv}{dt} = \frac{v^2}{x^2} \cdot \frac{dx}{dt} \]

### Step 5: Substitute the Known Values
Since the object is moving towards the mirror with speed \( V \), we have:
\[ \frac{dx}{dt} = -V \]
Substituting this into the equation:
\[ \frac{dv}{dt} = \frac{v^2}{x^2} \cdot (-V) \]

### Step 6: Analyze the Magnification
The magnification \( m \) for a convex mirror is given by:
\[ m = \frac{h'}{h} = -\frac{y}{x} \]
where \( h' \) is the height of the image and \( h \) is the height of the object. The magnification for a convex mirror is always less than 1, which implies:
\[ |m| < 1 \Rightarrow |y| < |x| \]

### Step 7: Conclusion on Speed of the Image
From the relationship derived, since \( |m| < 1 \), the speed of the image \( \frac{dv}{dt} \) will always be less than the speed of the object \( V \). Therefore, the speed of the image is always less than \( V \).

### Final Answer
The speed of the image is always less than \( V \).

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