Question

A person suffering from myopia was advised to wear corrective lens of power -2.5 D. At what distance should a student place an object from this lens so that it forms an image at a distance of 10 cm from the lens

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

Given :

• Power (P) = - 2.5 D
• Image Distance (v) = 10 cm

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To Find :

• Concave lens
• Object Distance (u)

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Solution :

We have formula for Focal length is :

$\huge{\boxed{\boxed{\rm{f \: = \: \dfrac{1}{P}}}}} \\ \\ \implies {\sf{f \: = \: \dfrac{1}{-2.5}}} \\ \\ \implies {\sf{f \: = \: - \: 0.4 \: m}} \\ \\ \implies {\sf{f \: = \: - \: 40 \: cm}} \\ \\ \leadsto {\boxed{\tt{Focal \: length \: = \: - \: 40 \: cm}}}$

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Now, use lens formula :

$\huge{\boxed{\boxed{\rm{\dfrac{1}{f} \: = \: \dfrac{1}{v} \: + \: \dfrac{1}{u}}}}} \\ \\ \implies {\sf{\dfrac{1}{u} \: = \: \dfrac{1}{f} \: - \: \dfrac{1}{v}}} \\ \\ \implies {\sf{\dfrac{1}{u} \: = \: \dfrac{-1}{40} \: - \: \dfrac{1}{10}}} \\ \\ \implies {\sf{\dfrac{1}{u} \: = \: \dfrac{-1 \: - \: 4}{40}}} \\ \\ \implies {\sf{\dfrac{1}{u} \: = \: \dfrac{-5}{40}}} \\ \\ \implies {\sf{\dfrac{1}{u} \: = \: \dfrac{1}{-8}}} \\ \\ \implies {\sf{u \: = \: - \: 8 \: cm}} \\ \\ \leadsto{\boxed{\tt{Object \: distance \: = \: -8 \: cm}}}$

Answer 2

${ \underline{\boxed{ \sf{ \huge{ \bold{ \orange{Answer}}}}}}} \\ \\ \star \sf \: \bold{GIVEN :} \\ \\ \longmapsto \sf \: power \: of \: lens \: (P) = - 2.5 \: D \\ \\ \longmapsto \sf \: position \: of \: image \: (v) = 10 \: cm \\ \\ \star \sf \: \bold{TO \: FIND :} \\ \\ \longmapsto \sf \: position \: of \: object \: (u) \\ \\ \star \sf \: \bold{FORMULA :} \\ \\ \longmapsto \: \sf \: First \: of \: all \: we \: have \: to \: find \: focal \: length \\ \sf \: \: \: \: \: \: \: \: \: \: \: to \: calculate \: position \: of \: object \\ \\ \longmapsto \: \sf relation \: between \: power \: of \: lens \: and \: focal \: length \\ \\ \leadsto \: \underline{\boxed{ \sf{ \red{ \bold{f = \frac{1}{P} }}}}} \\ \\ \longmapsto \sf \: lens \: formula \\ \\ \leadsto \sf \: \underline{\boxed{ \bold{ \red{ \frac{1}{u} + \frac{1}{v} = \frac{1}{f} }}}} \\ \\ \star \sf \: \bold{CALCULATION :} \\ \\ \longmapsto \sf \: f = \frac{1}{P} = \frac{1}{ - 2.5} = - 40 \: cm \\ \\ \longmapsto \sf \: \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \\ \\ \longmapsto \sf \: \frac{1}{u} = - \frac {1}{40} - \frac{1}{10} = - \frac{1}{8} \\ \\ \leadsto \sf \: \underline{\boxed{ \bold{ \blue{u = - 8 \: cm}}}}$