CBSE Class 12 Physics 2013 Paper

Question : 20 of 29

Marks: +1, -0

A small bulb (assumed to be a point source) is placed at the bottom of a tank containing water to a depth of

80 {\cm}

. Find out the area of the surface of water through which light from the bulb can emerge. Take the value of the refractive index of water to be

\;{4}/{3}

Solution:

Using Snell's law,

\;µ=\;{ sin \;i}/{ sin \;r}

\;\;\text " Or, "\; \;\; \;{4}/{3}=\;{ sin \;90^{∘}}/{ sin \;r}

\; ∴ \;\; i= sin \;^{-1} \;{3}/{4}=48.59^{∘}

\;\;\text " Now, "\; \;\; \tan i=\;{A B}/{O B}=\;{R}/{0.8}

\; ∴ \;\; R=0.8 × \tan 48.59^{∘}

\;=0.8 × 1.134=0.9 {\m}

Area of surface of water through which light will emerge

\;=π R^2=π ×(0.9)^2

\;=2.54 {\m}^2

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