# An object is placed at a distance of 10 cm to left on axis of a convex lens A of focal length 20 cm.Another 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 magnification of final image

$(a)\;1.5 \\ (b)\;1.33 \\ (c)\;2 \\ (d)\;1$

there for 1st lens $u_1= -10 \;cm$
$f_1= 20 cm$
$\large\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}$
=> $\large\frac{1}{v_1}=\frac{1}{20}-\frac{1}{10}$
$v_1=-20 \;cm$
ie the image is virtual and hence lies on the same side of object .
This will behave as an object for second lens
For second lens,
$\large\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}$
Here $u_2= -(20+5)$
$f_2= 10\;cm$
$\large\frac{1}{v_2}+\frac{1}{25}=\frac{1}{10}$
=> $v_2=\large\frac{50}{3}$
$\qquad= 16 \large\frac{2}{3}\;cm$
ie the final image is at a distance of $16 \large\frac{2}{3}\;cm$ on right of second lens.
Magnification of image
$m= \large\frac{v_1}{u_1} \frac{v_2}{u_2}$
$\quad= \large\frac{20}{10}. \frac{50}{3 \times 25}=\frac{4}{3}$
$\qquad= 1.33$
Hence b is the correct answer.