Question

A diverging lens has a focal length of 23.9 cm. An object 2.1 cm in height is placed 100 cm in front of the lens. Locate the position of the image. Answer in units of cm. 007 (part 2 of 3) 10.0 points What is the magnification? 008 (part 3 of 3) 10.0 points Find the height of the image. Answer in units of cm.

Answer 1

Explanation:

Given that,

Focal length of the diverging lens, f = -23.9 cm

Height of the object, h = 2.1 cm

Object distance, u = -100 cm

(a) Let v is the image distance. It can be calculated using lens formula as :

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

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

$\dfrac{1}{v}=\dfrac{1}{(-100)}+\dfrac{1}{(-23.9)}$

v = -19.28 cm

So, the image is located in from of the lens at a distance of 19.28 cm.

(b) The magnification of lens is given by :

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

$m=\dfrac{-19.28}{-100}$

m = 0.192

(c) Let h' is the height of the image. It can be calculated as :

$m=\dfrac{h'}{h}$

$h'=h\times m$

$h'=2.1\times 0.192$

h' = 0.403 cm

Hence, this is the required solution.