Question

A real $1mm$ long object is kept perpendicular to the principal axis of a concave mirror. The size of the image formed is $1/4mm$. Now the mirror starts moving, away from the object with a velocity $2cm/s$ along the principal axis, then

• A. the velocity of the image along the principle axis at the given instant will be $\frac{17}{8}$ cm/s towards the mirror.
• B. the length of the image will decrease as the mirror moves
• C. the speed of the image will always be more than 2 cm/s
• D. the length of the image will remain constant

Answer 1

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

The correct options are
A the velocity of the image along the principle axis at the given instant will be $\frac{17}{8}$ cm/s towards the mirror.
B the length of the image will decrease as the mirror moves
C the speed of the image will always be more than 2 cm/s

Lateral magnification $m=1/4$

With respect to mirror velocity of object $=2cm/s$ towards left.

Velocity of image $={\left(\frac{1}{4}\right)}^2\times 2cm/s$ towards right =$\frac{2}{16}=\frac{1}{8}$

With respect to ground velocity of $V_{image}$= $(\frac{1}{8}+2)\frac{cm}{s}=\frac{17}{8}$ towards right.

And as mirror starts moving, the length of image will decreases simultaneously.

Since, image formed is diminished the object has to be placed between $2f$ and infinity, now that the mirror is moving away from object the length of the image will decrease continuously.

$V_{image}-2=\frac{1}{16}\left(V_{object}\right)$
$\because V_{image}=2+\frac{1}{16}\left(V_{object}\right)$
So, velocity of image is always greater than $2m/s.$

As, the lenght of image is decreasing as the mirror moves hence, option (D) incorrect.
So, option $A$, $B$ and $C$ matches correctly.