Question

An object of height 2.2 cm is placed 27 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 11 cm from it, there is a converging lens of the same focal length.

Answer 1

Answer:

(a).The location of the final image beyond the converging lens is 67.2 cm.

(b). The magnification is 3.2

Explanation:

Given that,

Object height = 2.2 cm

Object distance = 27 cm

Focal length =16 cm

Distance = 11 cm

Suppose find the location of the final image, beyond the converging lens

We need to calculate the image distance

Using formula of lens

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

Put the value into the formula

$\dfrac{1}{-16}=\dfrac{1}{v}+\dfrac{1}{27}$

$\dfrac{1}{v}=\dfrac{1}{-16}-\dfrac{1}{27}$

$v=-10.0\ cm,$

We need to calculate the image distance from converging lens

Using formula of lens

Here, u = -10-11=-21 cm

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

Put the value into the formula

$\dfrac{1}{16}=\dfrac{1}{v}+\dfrac{1}{21}$

$\dfrac{1}{v}=\dfrac{1}{16}-\dfrac{1}{21}$

$v=67.2\ cm,$

(b). We need to calculate the magnification of the final image

Using formula of magnification

$m=\dfrac{v}{u}$

Put the value into the formula

$m=\dfrac{67.2}{-21}$

$m=-3.2\ cm$

The magnification is 3.2

Negative sign show the inverted image.

Hence, (a). The location of the final image beyond the converging lens is 67.2 cm.

(b). The magnification is 3.2