Question

(b) A concave lens has focal length of 20 cm. At what distance from the
lens a 5 cm tall object be placed so that it forms an image at 15 cm
from the lens? Also calculate the size of the image formed.

Answer 1

Given :

Focal length = 20cm

Height of object = 5cm

Distance of image = 15cm

Type of lens : concave

To Find :

■ Focal length of concave lens is taken negative and that of convex lens is taken positive.

Distance of object can be measured by using lens formula which is given by

$\bigstar\:\underline{\boxed{\bf{\red{\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}}}}}$

• u denotes distance of object
• v denotes distance of image
• f denotes focal length

By substituting the given values;

$:\implies\sf\:\dfrac{1}{-15}-\dfrac{1}{u}=\dfrac{1}{-20}$

$:\implies\sf\:\dfrac{1}{u}=-\dfrac{1}{15}+\dfrac{1}{20}$

$:\implies\sf\:\dfrac{1}{u}=\dfrac{-4+3}{60}$

$:\implies\sf\:\dfrac{1}{u}=\dfrac{-1}{60}$

$:\implies\:\underline{\boxed{\bf{\green{u=-60\ cm}}}}$

By applying formula magnification :

$:\implies\sf\:m=\dfrac{v}{u}=\dfrac{h_{image}}{h_{object}}$

$:\implies\sf\dfrac{(-15)}{(-60)}=\dfrac{h_i}{5}$

$:\implies\sf\dfrac{1}{4}=\dfrac{h_i}{5}$

$:\implies\sf\:h_i=\dfrac{5}{4}$

$:\implies\:\underline{\boxed{\bf{\pink{h_i=1.25\:cm}}}}$

Answer 2

Given :-

☄ Focal length, f = 20cm

☄ Height of object, h = 5cm

☄ Distance of image, u = 15cm

To Find :-

object distance and size of the image formed

Solution :-

by using len formula,

a formula which gives the relationship between image distance (v), object distance (u) and focal length (f) of the lens is known as the lens formula.

the lens formula can be written as :-

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

In concave lens focal length is considered negative

$\leadsto\sf{\dfrac{1}{-15}-\dfrac{1}{u}=\dfrac{1}{-20} }$

$\leadsto\sf{\dfrac{1}{u}=-\dfrac{1}{15}+\dfrac{1}{20}}$

$\leadsto\sf{\dfrac{1}{u}=\dfrac{-4+3}{60}}$

$\leadsto\sf{\dfrac{1}{u}=\dfrac{-1}{60}}$

$\leadsto{\sf{u=-60\ cm}}$

hence, object distance is -60cm.

now, using magnification formula .i.e.,

$\leadsto\bf magnification =\dfrac{v}{u}=\dfrac{h'}{h}$

$\leadsto\sf{\dfrac{-15}{-60}=\dfrac{h'}{5}}$

$\leadsto\sf{\dfrac{1}{4}=\dfrac{h'}{5}}$

$\leadsto\sf {h'=\dfrac{5}{4}}$

$\leadsto \sf{ h' = 1.25cm}$

thus, the size of the image is 1.25cm

$\:$

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