Question

An object 30cm tall is placed on the principal axis of convex lens and it's 60cm tall image is formed on the screen placed at a distance of 40cm from the object. What is the focal length of the lens?​

Answer 1

Given :

In convex lens,

Object height = 30cm

Image height = 60cm

Object distance = - 40cm

To find :

The focal length of the lens

Solution :

We know that,

» The ratio of image distance to the object distance is equal to the the ratio of image height to the image height

$\dashrightarrow \sf \dfrac{h'}{h} = \dfrac{v}{u}$

$\dashrightarrow \sf \dfrac{60}{30} = \dfrac{v}{ - 40}$

$\dashrightarrow \sf 2= \dfrac{v}{ - 40}$

$\dashrightarrow \sf v = - 80 \: cm$

Now, using lens formula that is,

» The formula which gives the relationship between image distance, object distance and focal length of a lens is known as the lens formula.

The lens formula can be written as :

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

where,

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

by substituting all the given values in the formula,

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

$\dashrightarrow \sf \dfrac{1}{( - 80)} - \dfrac{1}{( - 40)}= \dfrac{1}{f}$

$\dashrightarrow \sf - \dfrac{1}{ 80} + \dfrac{1}{40}= \dfrac{1}{f}$

$\dashrightarrow \sf \dfrac{ - 1 + 2}{ 80} = \dfrac{1}{f}$

$\dashrightarrow \sf \dfrac{1}{f} = \dfrac{1}{80}$

$\dashrightarrow \underline{ \boxed{\sf f = 80 \: cm}}$

Thus, the focal length of the lens is 80 cm