Question

A concave lens of focal length 60 cm is used to form an image of an object of length 9cm kept at a distance of 30 cm from it. Use lens formula to determine the nature, position, and length of the image formed. Also draw labelled ray diagram to show the image formation in the above case.​

Answer 1

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Concave lens-

focal length (f)=-60\; cm

Object length (h)=9\; cm

Object distance (u)=-30\; cm

Lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$

$v=\frac{-1}{60}+\left ( \frac{-1}{30} \right )$

$m=\frac{v}{u}=\frac{-20}{-30}=\frac{2}{3}$

$m=\frac{{h}'}{h}\Rightarrow {h}'=m\times h$

${h}'=\frac{2}{3}\times 9$

${h}'=6\; cm$

Image is virtual, erect and smaller than object.

Answer 2

hence, nature of image is virtual and diminished , image is placed between object and lens and length of image is 6cm