Question

Calculate the image distance when on object is place d at 15cm from a convex lens of focal length of 25cm

Answer 1

$\color{darkblue}\underline{\underline{\sf Given-}}$

➩ In Convex Lens

• Object placed at Distance (u) = -15 cm
• Focal length (f) = +25cm

$\color{darkblue}\underline{\underline{\sf To \; Find-}}$

• Image Distance (v).

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❏ Lens Formula

$\color{violet}\bullet\underline{\boxed{\sf \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}}$

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$\implies{\sf \dfrac{1}{25}=\dfrac{1}{v}-\left(-\dfrac{1}{15}\right) }$

$\implies{\sf \dfrac{1}{25}=\dfrac{1}{v}+\dfrac{1}{15} }$

$\implies{\sf \dfrac{1}{v}=\dfrac{1}{25}-\dfrac{1}{15}}$

$\implies{\sf \dfrac{1}{v}=\dfrac{15-25}{25×15} }$

$\implies{\sf \dfrac{1}{v}=\dfrac{-10}{375}}$

$\implies{\sf v=-\dfrac{375}{10} }$

$\color{red}\implies{\sf Image \: Distance (v)=-37.5cm}$

$\underline{\sf Nature \; Of \: Image -}$

• Object is placed between F and O
• Image is virtual , erect and enlarged .

Answer 2

$\mathtt{ \huge{ \fbox{Solution :)}}}$

By sign convention ,

• Image Distance (u) = - 15 cm
• Focal length (f) = 25 cm

We know that , the lens formula is given by

$\large \mathtt{ \fbox{ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} }}$

Thus ,

$\sf \hookrightarrow \frac{1}{25} = \frac{1}{v} - ( - \frac{1}{15} ) \\ \\ \sf \hookrightarrow \frac{1}{v} = \frac{1}{25} - \frac{1}{15} \\ \\ \sf \hookrightarrow \frac{1}{v} = \frac{15 - 25}{375} \\ \\ \sf \hookrightarrow v = - \frac{375}{10} \\ \\ \sf \hookrightarrow v = - 37.5 \: \: cm$

Here , the negative sign of v indicates the image is virtual

Hence , the image distance from pole is 37.5 cm

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