Question

A concave lens has focal length 15cm at what distance should the object from the lens to be placed so that it forms an image at 10cm from the lens also find the magnification produced by lens?

Answer 1

hope your will be clear
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thank you!

Answer 2

Answer:

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.

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

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

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

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

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

$\rule{150}2$

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

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

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

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