An object is placed 10cm from a converging lens of foal length 15cm. Calculate the magnification of the image formed

A.

3.0

B.

1.5

C.

0.6

D.

0.3

Correct answer is C

* produces a virtual image when object distance(u) is less than its focal length(f).

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

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

\(\frac{1}{15} + \frac{1}{10} = \frac{2 + 3}{30}\)

\(\frac{1}{v} = \frac{5}{30}\)

v = 6.0 cm 

but m = \(\frac{v}{u} = \frac{6}{10} = 0.6\)

Note that the negative sign only shows the placement of the image. It has no effect on the magnification basically.