Question

A particle moves towards a concave mirror of focal length $30\text{cm}$ along its axis and with a constant speed of $4\text{cm/sec}$.

• A. At the instant the particle is $90\text{cm}$ from the pole speed of image is $1\text{cm/sec}$
• B. At the instant the particle is $90\text{cm}$ from the pole speed of image w.r.t. particle is $5\text{cm/sec}$
• C. At the instant the particle is $90\text{cm}$ from the pole particle and image move towards each other
• D. The particle approaches pole, velocity of image always increases

Answer 1

Answer: (A), (B), (C)

The correct options are
A At the instant the particle is $90\text{cm}$ from the pole speed of image is $1\text{cm/sec}$
B At the instant the particle is $90\text{cm}$ from the pole speed of image w.r.t. particle is $5\text{cm/sec}$
C At the instant the particle is $90\text{cm}$ from the pole particle and image move towards each other

Magnification $m=\frac{f}{f-u}=\frac{-30}{-30+90}=\frac{-1}{2}$

(A) Let $V_{I/m}$ be the velocity of image with respect to the mirror and $V_{O/m}$ be the velocity of object with respect to the mirror. Then,
$V_{I/m}=-m^2{V}_{O/m}=-\frac{1}{4}\times 4=-1\text{cm/s}$

(B) The velocity of image with respect to the object is given by$V_{I/O}=V_I-V_O=-1-\left(4\right)=-5\text{cm/s}$

(C) $V_{I/m}=-m^2{V}_{O/m}$
and since the mirror is stationary, so particle and image move towards each other.
Also since speeds are asked in the options we can ignore the minus signs in the final answers.