Question

A 5.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10cm. The distance of the object from the lens is 25 cm. Find the nature, position & size of the image. Also find its magnification​

Answer 1

Given :

In convex lens,

Object height : 5 cm.

Object distance : 25 cm.

Focal length : 10 cm.

To find :

The nature, position, size of the mirror and it's magnification.

Solution :

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,

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

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

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

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

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{5 - 2}{50}$

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

$\dashrightarrow \sf v= \dfrac{50}{3}$

$\dashrightarrow \sf v= 16.6 \: cm$

Thus, the position of the image is 16.6 cm.

we know that,

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

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

$\dashrightarrow \sf \dfrac{h'}{5} = \dfrac{16.6}{25}$

$\dashrightarrow \sf h' = \dfrac{16.6 \times 5}{25}$

$\dashrightarrow \sf h' = \dfrac{16.6}{5}$

$\dashrightarrow \sf h' = 3.3 \: cm$

Thus, the height of the image is 3.3 cm.

we know that,

$\dashrightarrow \sf m = \dfrac{v}{u}$

$\dashrightarrow \sf m = \dfrac{16.6}{25}$

$\dashrightarrow \sf m = 0.6$

Thus, the magnification of the image is 0.6

Nature of the image :

• The image is erect and virtual.
• The image is diminished.