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Q. A beam of light converges at a point P. A concave lens of focal length 16 cm is placed in the path of this beam 12 cm from P. The location of the point, from the lens, at which the beam would now converge is

MGIMS WardhaMGIMS Wardha 2014

Solution:

Here, $ u=+12cm,f=-16\,cm=v=? $ By lens equation $ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} $ Substituting the values $ \frac{1}{-16}=\frac{1}{v}-\frac{1}{12} $ $ \frac{1}{v}=\frac{1}{12}-\frac{1}{16}=\frac{4-3}{48}=\frac{1}{48} $ $ v=+48\,cm $ The image of virtual object P forms at P at a distance of 48 cm from lens.