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A point object is placed at equal distance $3 f$ in front of a concave mirror, a convex mirror and a plane mirror seperately (event-1). Now, the distance is decreased to $1.5 f$ from all the three mirrors (event-2). Magnitude of focal length of convex mirror and concave mirror is $f$ , Then, choose the correct option.

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Maximum distance of image in event-1 from the mirror is from plane mirror.

Minimum distance of image in event-1 from the mirror is from convex mirror.

Maximum distance of image in event-2 from the mirror is from concave mirror.

Minimum distance of image in event-2 from the mirror is from plane mirror.

answer is A, B, C.

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Detailed Solution

In event-1:

For concave mirror,

$\frac{1}{v} + \frac{1}{- 3 f} = \frac{1}{- f} \Rightarrow v = - 1.5 f$

For convex mirror,

$\frac{1}{v} + \frac{1}{- 3 f} = \frac{1}{+ f} \Rightarrow v = \frac{3}{4} f = 0.75 f$

For plane mirror,

$v = + 3 f$

In event-2:

For concave mirror,

$\frac{1}{v} + \frac{1}{- 1.5 f} = \frac{1}{- f} \Rightarrow v = - 3 f$

For convex mirror,

$\frac{1}{v} + \frac{1}{- 1.5 f} = \frac{1}{+ f} \Rightarrow v = \frac{3}{5} f = 0.6 f$

For plane mirror,

$v = + 1.5 f$

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