Question

A 10cm tall object is placed perpendicular to the principle axis of convex lens. focal length is 30cm and object distance is 20cm.
find nature , size , position of image.​

Answer 1

Given :

Object height, h = 10cm.

focal length, f = 30cm

object distance, u = -20cm.

To find :

Nature, size and position of image.

Solution :

using lens formula .i.e.,

A 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 :-

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

${ \leadsto{ \bf{ \dfrac{1}{v} - \dfrac{1}{( - 20)} = \dfrac{1}{30} }}}$

${ \leadsto{ \bf{ \dfrac{1}{v} + \dfrac{1}{20} = \dfrac{1}{30} }}}$

${ \leadsto{ \bf{ \dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{30} }}}$

${ \leadsto{ \bf{ \dfrac{1}{v} = \dfrac{2 - 3}{60} }}}$

${ \leadsto{ \bf{ \dfrac{1}{v} = - \dfrac{1}{60} }}}$

${ \leadsto{ \bf{v = - 60 \: cm }}}$

thus, image distance = -60 cm.

Magnification :

The Linear magnification produced by a lens is equal to the ratio of image distance to the object distance .i.e.,

${ \leadsto{ \bf{ m = \dfrac{v}{u} }}}$

${ \leadsto{ \bf{ m = \dfrac{ - 60}{ - 20}}}}$

${ \leadsto{ \bf{ m = 3}}}$

The Linear magnification is the ratio of the height of the image to the height of the object.

${ \leadsto{ \bf{ m = \dfrac{h'}{h} }}}$

${ \leadsto{ \bf{ 3 = \dfrac{h'}{10} }}}$

${ \leadsto{ \bf{ h' = 30 \: cm}}}$

thus, height of image = 30 cm.

Nature of the image :

• the image is real and inverted
• the image is enlarged