Question

An object of height 5 cm is placed at a distance of 40cm from a convex lens of focal length 50cm. Calculate the size, position and nature of image formed

Answer 1

Given,

Object height O = 5cm

Image height I = ?

Object distance u = 40 cm

Focal length f = 50cm

Image distance v = ?

$\huge \boxed{\frac{1}{v} - \frac{1}{u}= \frac{1}{f}}$

$\huge \boxed{\frac{1}{v} = \frac{1}{f} + \frac{1}{u} }$

Since its convex lens,
object distance will be negative.

$\frac{1}{v} = \frac{1}{50} + \frac{1}{-40}$

$\frac{1}{v} = \frac{1}{50} - \frac{1}{40}$

$\frac{1}{v} = \frac{4- 5}{200}$

$\frac{1}{v} = \frac{-1}{200}$

$\boxed{v = -200cm}$

We know,

$\boxed{MAGNIFICATION = \frac{v}{u } = \frac{I}{O}}$

$MAGNIFICATION = \frac{20\cancel{0}}{4\cancel{0}} = \frac{I}{5}$

$\boxed{I = 25cm }$

$\boxed{MAGNIFICATION = \frac{I}{O}}$

$MAGNIFICATION = \frac{25}{5}$

$MAGNIFICATION =5cm$

NATURE OF IMAGE

▶Since, magnification is more than 5 therefore image is magnified.

▶Since, image distance is negative image is formed on same side as od object.

▶Its a virtual image.

▶Convex lens only forms virtual image when object is between focus and optical center.

Answer 2

Step-by-step explanation:

the magnification is 5cm