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Q. A point source is located $2.75 \,cm$ below the surface of a lake. The area of surface that transmits all the light that emerges from the surface is :

Haryana PMTHaryana PMT 1999

Solution:

Now, using the relation
$ r=\frac{h}{\sqrt{{{\mu }^{2}}-1}} $
or $ A=\pi {{r}^{2}}=\frac{\pi {{h}^{2}}}{{{\mu }^{2}}-1} $
$ =\frac{3.14\times {{(2.75)}^{2}}}{{{\left( \frac{4}{3} \right)}^{2}}-1}=30.4{{m}^{2}} $