Physics, asked by alinajay159, 11 months ago

An object 2.5 cm high is placed 6.0 cm from a converging lens with a focal length of 4.0cm. Determine (a) the image distance, (b) the image size and (c) the magnification.

Answers

Answered by Anonymous

Answer :

(a) The image distance is 12cm

(b) The image size is 5cm

Given :

An object of 2.5cm height is 6.0cm from a converging lens
A focal length of the lens is 4.0cm

To Find :

The image distance
The image size
The magnification

Formulae used :

Lens formula

$\sf \star \: \: \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Formula for magnification of the lens

$\sf \star \: \: m = \dfrac{h_{2}}{h_{1}} =\dfrac{v}{u}$

Solution :

Given ,

Object distance , u = -6cm

Height of the object , h1 = 2.5cm

Focal length , f = 4cm

Applying mirror formula ,

$\sf \implies \dfrac{1}{v}-\dfrac{1}{-6}=\dfrac{1}{4}\\\\ \sf \implies \dfrac{1}{v}+\dfrac{1}{6}=\dfrac{1}{4}\\\\ \sf \implies \dfrac{1}{v}=\dfrac{1}{4} - \dfrac{1}{6}\\\\ \sf \implies \dfrac{1}{v} =\dfrac{1}{12}\\\\ \sf \implies v = 12cm$

Therefore , the image distance is 12cm

Now magnification ,

$\sf \implies m = \dfrac{v}{u} \\\\ \sf \implies m = \dfrac{12}{-6}\\\\ \sf \implies m = -2$

Thus ,the magnification is -2

The negative sign represents image formed is real and inverted

Again we have ,

$\sf \implies m = \dfrac{h_{2}}{h_{1}}\\\\ \sf \implies -2 = \dfrac{h_{2}}{2.5cm } \\\\ \sf \implies h_{2} = -5cm$

The size of the image is 5cm

Attachments:

Answered by Anonymous

Object distance , u = -6cm

Height of the object , h1 = 2.5cm

Focal length (f) = 4cm

Using mirror formula:-

$⇢\dfrac{1}{v}=\dfrac{1}{4} - \dfrac{1}{6}\\\\ ⇢ \dfrac{1}{v} =\dfrac{1}{12}\\\\ ⇢ v = 12cm$

Therefore , the image distance is 12cm

Magnification formula:-

$m = \dfrac{v}{u} \\\\ ⇢m = \dfrac{12}{-6}\\\\ ⇢m = -2$

Thus ,the magnification is -2

The negative sign represents image formed is real and inverted

Again we have,

$m = \dfrac{h_{2}}{h_{1}}\\\\ ⇢ -2 = \dfrac{h_{2}}{2.5cm } \\\\ ⇢h_{2} = -5cm$