An object is placed at a distance of 10 cm to the left on the axis of a convex lens A of focal length 20 cm. A second convex lens of focal length 10 cm is placed co-axially to the right of the lens A at a distance of 5 cm from A. Find the position of the final image and its magnification. Trace the path of the rays.

1.33

Here for 1st lens, $u_1 = -10 cm$

$f_1 = 20 cm$

$\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}$

$⇒\frac{1}{v_1}=\frac{1}{20}-\frac{1}{10}$

$⇒v_1=-20cm$

i.e. the image is virtual and hence lies on the same side of the the object. This will behave as an object for the second lens.

For 2nd lens

$\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}$

Here $u_2 = -(20 + 5)$

$f_2 = 10 cm$

$\frac{1}{v_2}+\frac{1}{25}=\frac{1}{10}⇒v_2=\frac{50}{3}=16\frac{2}{3}cm$

i.e. the final image is at a distance of $16\frac{2}{3}$ 16 cm on the right of the second lens.

The magnification of the image is given by, $m=\frac{v_1}{u_1}.\frac{v_2}{u_2}=\frac{20}{10}.\frac{50}{3×25}=\frac{4}{3}=1.33$