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Q. A needle placed $45 cm$ from a lens forms an image on the screen placed $90 cm$ on the other side of the lens. What is the size of image if the size of needle is $5 cm ?$ And what is the focal length of lens?

Ray Optics and Optical Instruments

Solution:

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$ \Rightarrow \frac{1}{90}+\frac{1}{45}$
$\frac{1}{f}=\frac{3}{90}$
$f=30 cm$
$\frac{I}{O}=\frac{-v}{u}$
$\frac{I}{5}=\frac{-90}{-45} $
$\Rightarrow I=10$ cm