Question

A concave lens has focal length of 120m. At what ustalice should the object
from the lens be placed so that it
forms an image at 8cm from the lens? Also,
find the magnification produced by the lens,​

Answer 1

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• Focal length, f = 120 cm

•Image distance, v = 8cm

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• Object distance, u

• Magnification, m

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

$\underline{\:\textsf{ As we know the lens formula :}}$

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

$\underline{\:\textsf{ Putting the values :}}$

$\dashrightarrow \sf \dfrac{-1}{120}= \dfrac{1}{8}-\dfrac{1}{u}\\ \\ \dashrightarrow \sf \dfrac{1}{u}= \dfrac{1}{8}+\dfrac{1}{120}\\ \\ \dashrightarrow \sf \dfrac{1}{u}= \dfrac{15+1}{120}\\ \\\dashrightarrow \sf \dfrac{1}{u}= \dfrac{16}{120}\\ \\\dashrightarrow \sf u = \dfrac{15}{2} cm \ or \ 7.5 cm$

ㅤ$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ Object distance </p><p>\textbf{ 7.5 cm }}}.$

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Now find the magnification.

$\underline{\:\textsf{ We know that :}}$

$\boxed{\sf m= \dfrac{(v) \ Image \ distance }{(u)\ object \ distance } }$

$\underline{\:\textsf{ Putting the values :}}$

$\dashrightarrow \sf m = \dfrac{8}{\dfrac{15}{2}}\\ \\ \dashrightarrow \sf m= 8\times \dfrac{2}{15}\\ \\ \dashrightarrow \sf m= \dfrac{8\times 2}{15}\\ \\ \dashrightarrow \sf m = \dfrac{16}{15}$

ㅤ$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ Magnification </p><p>\textbf{ 16/15 }}}.$

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Answer 2

Answer:

$16 \div 5 \: is \: th \:e \: answer \: nut \: you \: want \: to \: write \: it \: in \: fraction$

Explanation:

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