A lens of focal length 15cm forms an erect image which is three times the...

A lens of focal length 15cm forms an erect image which is three times the size of the object. The distance between the object and the image is ___.

F= 15cm

M = 3

M = \(\frac{v}{u}\)

V =3u

\(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)

\(\frac{1}{15}\) = \(\frac{1}{u}\) - \(\frac{1}{3u}\)

\(\frac{1}{15}\) = \(\frac{2}{3u}\)

U =10cm

D = u + v

= 10 +30 = 40cm

There is an explanation video available below.

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