Question

An object of height 3.0 cm is placed at 25 cm in front of a diverging lens of focal length 20 cm. Behind the diverging lens, there is a converging lens of focal length 20 cm. The distance between the lenses is 5.0 cm. Find the location and size of the final image.

di = _____ cm

hi = _____ cm

Answer 1

Height of image, $h=3.0 \space cm$

Image distance, $u_1=-25\space cm$

Focus, $f_1=-20\space cm$

Let distance of image formed by the diverging lens $=v_1$

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

$\dfrac{1}{v_1}=-\dfrac{1}{20}-\dfrac{1}{25}$

$v_1=-\dfrac{100}{9}\space cm$

Height of image, $h_1'=h\times\dfrac{v_1}{u_1}$

$h_1'=\dfrac{4}{3}\space cm$

The image formed by the diverging lens will be the object for converging lens

Distance of object for converging lens, $u_2=-(v_1+5)$

$u_2=-\dfrac{135}{9}\space cm$

Focus, $f_2=20\space cm$

Image distance $=v_2$

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

$\dfrac{1}{v_2}=\dfrac{1}{20}-\dfrac{9}{135}$

$v_2=-60\space cm$

Height of the image formed, $h_2'=h_1'\times\dfrac{v_2}{u_2}$

$h_2'=\dfrac{16}{3}=5.33 \space cm$

Therefore distance of final image, di = 60 cm (left to the converging lens)

Height of final image, hi = 5.33 cm