Question

A converging lens (f1 = 12.5 cm) is located 28.8 cm to the left of a diverging lens (f2 = -6.18 cm). A postage stamp is placed 36.9 cm to the left of the converging lens. Locate the final image of the stamp relative to the diverging lens.

Answer 1

Answer:

The final image relative to the diverging lens at 9.0 cm.

Explanation:

Given that,

Focal length of diverging lens = -6.18 cm

Focal length of converging lens = 12.5 cm

Distance of object = 36.9 cm

We need to calculate the image distance of converging lens

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{12.5}-\dfrac{1}{-36.9}$

$\dfrac{1}{v}=\dfrac{988}{9225}$

$v=9.33\ cm$

We need to calculate the image distance of diverging lens

Now, object distance is

$u=28.8-9.33=19.47 cm$

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{-6.18}-\dfrac{1}{-19.47}$

$\dfrac{1}{v}=-\dfrac{22150}{200541}$

$v=-9.0\ cm$

The image is formed 9.0 cm to the left of the  diverging lens.

Hence, The final image relative to the diverging lens at 9.0 cm.