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Q.
lf the focal length of the lens is $20\, cm$. what is the distance of the image from the lens in the following figure?
BHUBHU 2006Ray Optics and Optical Instruments
Solution:
According to figure, the point on the right side of the lens of which rays converge will behave as virtual object at the lens.
$\therefore u = + 12\, cm, f = 20\, cm$
By lens formula
$ \frac{1}{f}= \frac{1}{v}- \frac{1}{u}$
$\therefore \frac{1}{20}= \frac{1}{v}- \frac{1}{12}$
or $ \frac{1}{v}=\frac{1}{20} + \frac{1}{12} = \frac{3+5}{60} = \frac{8}{60}$
or $ u = \frac{60}{8} = 7.5\, cm$
So, image will be formed on same side of the virtual object at a distance of 7.5 cm from the lens.
