Question

A concave lens has focal length of 15cm .At what distance should
the object from the lens be placed ,So that it forms an image at 10cm
from the lens?Also find the magnification produced by the lens.

Answer 1

Answer:

object should be placed 6 CM in front of lens

Explanation:

f=-15cm

v=-10cm

by mirror formula

1/f=1v-1/u

-1/15=1/10-1/u

1/u=-1/15-1/10

1/u=-1/6

u=-6cm

magnification=v/u=-10/-6=5/3

plz mark as brainliest

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.

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

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

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

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

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

$\rule{150}2$

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

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

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

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