A bright point object is kept at some distance from a lens of focal length 20 cm. If the object distance is changed $$\frac{5}{6}$$ times, the distance of the screen from the lens has to be changed by $$\frac{5}{3}$$ times to obtain a clear image on the screen. Find the distance (in cm) through which the screen has been moved.

For a real image with distances taken positive, $$\frac{1}{u} + \frac{1}{v} = \frac{1}{20}$$ and $$\frac{6}{5u} + \frac{3}{5v} = \frac{1}{20}$$, i.e. $$\frac{6}{u} + \frac{3}{v} = \frac{1}{4}$$. Subtracting three times the first equation gives $$\frac{3}{u} = \frac{1}{4} - \frac{3}{20} = \frac{1}{10}$$, so $$u = 30\text{ cm}$$ and $$v = 60\text{ cm}$$. The new screen distance is $$\frac{5}{3} \times 60 = 100\text{ cm}$$, so the screen has moved $$100 - 60 = 40\text{ cm}$$.