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

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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.

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