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A concave mirror of focal length 20cm is cut into two parts from the middle and the two parts are moved perpendicularly by a distance 1cm from the previous principle axis AB. The distance between the images formed by the two parts, if the object is kept at a distance of 10cm from the pole of the original principal axis is

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4 cm

2 cm

6 cm

3 cm

answer is A.

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Detailed Solution

u = - 10 cm, f = - 20 cm, v = ?

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

$\frac{- 1}{20} = \frac{- 1}{10} + \frac{1}{v}$

$\frac{1}{v} = \frac{1}{10} - \frac{1}{20}$

$\frac{1}{v} = \frac{1}{20}$

$v = 20 cm$

$m = \frac{v}{u} = \frac{- 20}{- 10} = 2 cm$

Two images formed are 2cm height from the respective principle axis one below the principle axis and other above principle axis.

$∴$ Distance between two images is 2 cm

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