The object distance u = 60cm

height of the real, inverted image \(\displaystyle{h}_{{{i}}}=-{2}{m}{m}\)

height of object \(\displaystyle{h}_{{{0}}}={4}{m}{m}\)

Thus, the magnification \(\displaystyle{M}=\frac{{h}_{{{i}}}}{{h}_{{{0}}}}=-\frac{{v}}{{u}}\)

- 2 / 4 = - v / 60

calculating we get v = 30cm

The focal lenghtof the lens is calculated using

1 / f = 1 / u + 1 / v

= 1 / 60 + 1 / 30

Thus, f = 20cms

height of the real, inverted image \(\displaystyle{h}_{{{i}}}=-{2}{m}{m}\)

height of object \(\displaystyle{h}_{{{0}}}={4}{m}{m}\)

Thus, the magnification \(\displaystyle{M}=\frac{{h}_{{{i}}}}{{h}_{{{0}}}}=-\frac{{v}}{{u}}\)

- 2 / 4 = - v / 60

calculating we get v = 30cm

The focal lenghtof the lens is calculated using

1 / f = 1 / u + 1 / v

= 1 / 60 + 1 / 30

Thus, f = 20cms