Question

ige formed.
20. When an object is placed 20 cm from a concave mirror, a real image magnified three times is formed. Find
(a) the focal length of the mirror.
(b) Where must the object be placed to give a virtual image three times the height of the object?
​physics question

Answer 1

Explanation:

Solution:-

$\sf (a) \: Object \: distance (u) = -20cm$

$\sf Magnification (m) = -3$

$\sf Image\: distance (v) = ?$

$\sf As, \: Magnification = \dfrac{-v}{u}$

$\dashrightarrow \sf -3 = \dfrac{-v}{-20}$

$\dashrightarrow \sf - v = - 20 \times -3$

$\dashrightarrow \sf - v = 60$

$\dashrightarrow \sf v = -60cm$

$\sf \therefore Image \: distance \: (v) = -60cm$

$\sf \underline{Using\: mirror \: formula}$

$\sf \dashrightarrow \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

$\sf \dashrightarrow \dfrac{1}{f} = \dfrac{1}{-60} + \dfrac{1}{-20}$

$\sf \dashrightarrow \dfrac{1}{f} = \dfrac{-1 - 3}{60} = \dfrac{-4}{60}$

$\sf \dashrightarrow \dfrac{1}{f} = \dfrac{-1}{15}$

$\sf \dashrightarrow f = -15cm$

$\therefore \sf \underline{Focal\: length = -15cm}$

$\sf (b) \: Magnification \: (m)= 3$

$\dashrightarrow \sf m = \dfrac{-v}{u}$

$\dashrightarrow \sf 3 = \dfrac{-v}{u}$

$\dashrightarrow \sf v = -3u$

$\sf \underline{Using\: mirror \: formula}$

$\sf \dashrightarrow \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

$\sf \dashrightarrow \dfrac{1}{-15} = \dfrac{1}{-3u} + \dfrac{1}{u}$

$\sf \dashrightarrow \dfrac{1}{-15} = \dfrac{-1 + 3}{3u}$

$\sf \dashrightarrow -15 = \dfrac{2}{3u}$

$\sf \dashrightarrow u = \dfrac{2 \times -15 }{3}$

$\sf \dashrightarrow u = -10cm$

Therefore, object should be placed 10cm infront of the mirror to give a virtual image three times the height of the object.