Question

An object 5 cm in length is held 25cm away from a converging lens of focal length 10cm .Draw the ray diagram and find the position, size and nature of the image formed?​

Answer 1

Given :

• Height of an object, h = 5cm.
• Position of object, u = - 25cm away from converging lens.
• Focal length, f = 10cm.

To find :

• Draw the ray diagram.
• Find the position.
• What is the size and nature of the image formed ?

Solution :

• Draw a ray diagram.

=> Ray diagram is provided in the attachment.

• Find the position.

Using the formula,

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

$\implies \sf \dfrac {1}{v} \ + \ \dfrac {1}{25} \ = \ \dfrac {1}{10}$

$\implies \sf \dfrac {1}{v} \ = \ \dfrac {1}{10} \ - \dfrac {1}{25}$

$\implies \sf \dfrac {1}{v} \ = \ \dfrac {3}{50}$

$\implies \sf \dfrac {1}{v} \ = \ \dfrac {50}{3}$

$\implies \sf 16.66 \ cm$

$\therefore$ Distance of an image is 16.66 cm.

• Size and nature of the image formed.

$\sf Magnification \ = \ \dfrac {v}{u}$

$\implies \sf magnification, m \ = \ \dfrac {16.66}{-25}$

$\sf magnification, m \ = \ -0.66$

Now,

$\implies \sf -0.66 \ = \ \dfrac {Height \ of \ image}{5 cm}$

$\sf Height \ of \ an \ image \ = \ 3.3 cm$

★ The negative sign of the image height represents that image formed is inverted.

★ Position of image = 1.66 cm at opposite side of lens.

★ Nature of the image is inverted and real.