Question

an object 3cm high is placed at a distance of 20cm from a convex lens of focal length 40 cm find the poistion nature size of the image​

Answer 1

$\Large\underline{\underline{\sf \green{Given}:}}$

• Object is placed at distance (u) = -20cm

• Focal lenght is at = -40cm

$\Large\underline{\underline{\sf \green{To\:Find}:}}$

• Position nature and size of image .

$\Large\underline{\underline{\sf \green{Formula\:Used}:}}$

Len's Formula

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

Here , v = positon of image .

$\Large\underline{\underline{\sf \green{Solution}:}}$

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

$\implies{\sf - \dfrac{1}{40}=\dfrac{1}{v}-\dfrac{1}{-20}}$

$\implies{\sf \dfrac{1}{v}=\dfrac{1}{-40}-\dfrac{1}{20}}$

$\implies{\sf v=\dfrac{-60}{800} }$

$\implies{\sf v= - \dfrac{800}{60} }$

$\implies{\sf v=-13.3cm }$

here , negative sign indicates that positon of image is on left side of lens.

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Nature of Image

• Virtual image (in front of lens )

• Erected and enlarged