An object is placed (a) 20 cm, (b) 4 cm, in front of a concave mirror of focal length 12 cm. Find the nature and position of the image formed in each case.

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Solution

(a)u=-20cm, f=-12cm
$1 over v plus 1 over u equals 1 over f fraction numerator begin display style 1 end style over denominator begin display style v end style end fraction plus fraction numerator begin display style 1 end style over denominator begin display style negative 20 end style end fraction equals fraction numerator begin display style 1 end style over denominator begin display style negative 12 end style end fraction fraction numerator begin display style 1 end style over denominator begin display style v end style end fraction equals fraction numerator begin display style negative 1 end style over denominator begin display style 12 end style end fraction plus fraction numerator begin display style 1 end style over denominator begin display style 20 end style end fraction equals fraction numerator negative 1 over denominator 30 end fraction v equals negative 30 c m$

Magnification, $m = \frac{- v}{u} = \frac{30}{- 20} = - 1.5$

So, the image is formed at a distance of 30 cm in front of the mirror which is real, inverted and magnified.

(b)u = -4cm, f = -12cm
$1 over v plus 1 over u equals 1 over f fraction numerator begin display style 1 end style over denominator begin display style v end style end fraction plus fraction numerator begin display style 1 end style over denominator begin display style negative 4 end style end fraction equals fraction numerator begin display style 1 end style over denominator begin display style negative 12 end style end fraction fraction numerator begin display style 1 end style over denominator begin display style v end style end fraction equals fraction numerator begin display style negative 1 end style over denominator begin display style 12 end style end fraction plus fraction numerator begin display style 1 end style over denominator begin display style 4 end style end fraction equals 2 over 12 v equals 6 space c m$

Magnification, $m = \frac{- v}{u} = \frac{- 6}{- 4} = 1.5$

The image is formed at a distance of 6 cm behind the mirror which is virtual, erect and magnified.

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