Question

An object is placed at a distance of 30 cm of a converging lens. The focal length of the lens is 40 cm. Find the distance of the image formed from the lens and the magnificent of the image.​

Answer 1

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

Given :

• Object Distance (u) = - 30 cm
• Focal Length (f) = 40 cm
• Convex Lens

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To Find :

• Distance
• Magnification
• Nature

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Solution :

We know that lens formula is :

$\large{\boxed{\sf{\dfrac{1}{f} \: = \: \dfrac{1}{v} \: - \: \dfrac{1}{u}}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{f} \: + \: \dfrac{1}{u}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{40} \: + \: \dfrac{-1}{30}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{3 \: - \: 4}{120}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{-1}{120}}} \\ \\ \implies {\sf{v \: = \: - \: 120 \: cm}} \\ \\ \underline{\sf{\therefore \: Image \:distance \: is \: - \: 120 \: cm}}$

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And we know that formula for magnification is :

$\large{\boxed{\sf{m \: = \: \dfrac{v}{u}}}} \\ \\ \implies {\sf{m \: = \: \dfrac{-120}{-30}}} \\ \\ \implies {\sf{m \: = \: 4}} \\ \\ \underline{\sf{\therefore \: Magnification \: is \: + \: 4}}$

• Nature of image is Virtual and Erect