Question

A concave lens has a focal length 15cm . At what distance should the object be placed from the lens so that it forms an image at 10cm from the lens ? What will be the nature and magnification of the image ?

Answer 1

AnswEr:

$\large\bold{\underline{\sf{\red{\:\:Given\:\:}}}}$

• f = -15cm [focal length of the lens]

• v = -10cm [Object Distance]

$\large\bold{\underline{\sf{\red{Using\: Lens\: Formula}}}}$

$\dag\:\:\large{\underline{\boxed{\sf{\blue{\dfrac{1}{f} \:=\:\dfrac{1}{v}\:-\:\dfrac{1}{u}}}}}}$

Concave lens always former virtual Image so, v is negative & f is also negative.

$\longrightarrow\sf\:\dfrac{\:\:1}{-15} = \dfrac{\:\:1}{-10} - \dfrac{1}{u}$

$\longrightarrow\sf\: \dfrac{1}{u} = \dfrac{\:\;1}{-10} - \dfrac{\:\:1}{-15}$

$\longrightarrow\sf\: \dfrac{1}{u} = \dfrac{-3\:+\:2}{30}$

$\longrightarrow\sf\: \dfrac{1}{u} = \dfrac{-1}{30}$

$\longrightarrow\large{\underline{\boxed{\sf{\blue{u\:= \: -30\:cm}}}}}$

$\rule{150}2$

$\diamond\:\:\bold{\underline{\sf{\red{Now\: Magnification}}}}$

$\longrightarrow\sf\: \dfrac{v}{u} = \dfrac{-10}{-30}$

$\longrightarrow\sf\: \dfrac{v}{u} = \dfrac{1}{3}$

$\longrightarrow\large{\underline{\boxed{\sf{\red{m \:=\:0.3}}}}}$

Answer 2

AnswEr:

$\large\bold{\underline{\sf{\red{\:\:Given\:\:}}}}$

• f = -15cm [focal length of the lens]

• v = -10cm [Object Distance]

$\large\bold{\underline{\sf{\red{Using\: Lens\: Formula}}}}$

$\dag\:\:\large{\underline{\boxed{\sf{\blue{\dfrac{1}{f} \:=\:\dfrac{1}{v}\:-\:\dfrac{1}{u}}}}}}$

Concave lens always former virtual Image so, v is negative & f is also negative.

$\longrightarrow\sf\:\dfrac{\:\:1}{-15} = \dfrac{\:\:1}{-10} - \dfrac{1}{u}$

$\longrightarrow\sf\: \dfrac{1}{u} = \dfrac{\:\;1}{-10} - \dfrac{\:\:1}{-15}$

$\longrightarrow\sf\: \dfrac{1}{u} = \dfrac{-3\:+\:2}{30}$

$\longrightarrow\sf\: \dfrac{1}{u} = \dfrac{-1}{30}$

$\longrightarrow\large{\underline{\boxed{\sf{\blue{u\:= \: -30\:cm}}}}}$

____________________________

$\diamond\:\:\bold{\underline{\sf{\red{Now\: Magnification}}}}$

$\longrightarrow\sf\: \dfrac{v}{u} = \dfrac{-10}{-30}$

$\longrightarrow\sf\: \dfrac{v}{u} = \dfrac{1}{3}$

$\longrightarrow\large{\underline{\boxed{\sf{\red{m \:=\:0.3}}}}}$