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We know,
1/f = 1/v + 1/u
=> 1/v = 1/f – 1/u
=> 1/v = (u – f)/(uf)
=> v = uf/(u – f)
Now,
dv/dt = [f/(u – f)]du/dt + uf(-1)(u – f)-2du/dt
=> dv/dt = (du/dt)[{f(u – f) – uf}/(u – f)2]
=> dv/dt = -(du/dt)[(f2)/(u – f)2]
A/Q,
Object distance is, u = -5 m
Object’s speed of approaching, du/dt = -20 m/s
Focal length, f = 30 cm = -0.3 m
So,
dv/dt = -(-20)[(0.32)/(-20 + 0.3)2]
=> dv/dt = 0.0046 m/s
=> dv/dt = 4.6 mm/s
This is the speed of the image.
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