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

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

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

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

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

answer is A, B, C.

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Detailed Solution

In event -1
For concave mirror, $\frac{1}{v} + \frac{1}{- 3 f} = \frac{1}{- f}$

$∴ v = - 1.5 f$

For convex mirror,

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

For plane mirror, $v = + 3 f$

Inevent-2
For concave mirror,

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

$∴ v = - 3 f$

For convex mirror,

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

For plane mirror, $v = + 1.5 f$

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