A) concave lens
B) convex lens
C) Neither convex nor concave
D) Cannot say
Correct Answer: A
Solution :
The focal length of air lens is \[\frac{1}{f}=\frac{{{\mu }_{2}}-{{\mu }_{1}}}{{{\mu }_{1}}}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] where\[{{\mu }_{2}}\] is refractive index of water and\[{{\mu }_{1}}\]is refractive index of air bubble. \[{{R}_{1}}\]and \[{{R}_{2}}\] are radii of curvatures of air bubble. Here, \[{{\mu }_{2}}=1\] \[{{\mu }_{1}}=\frac{4}{3}\] \[{{R}_{1}}=R\] \[{{R}_{2}}=-R\] \[\therefore \] \[\frac{1}{f}=\frac{1-4/3}{4/3}\left\{ \frac{1}{R}-\left( -\frac{1}{R} \right) \right\}\] or \[\frac{1}{f}=\frac{-\frac{1}{3}}{\frac{4}{3}}\times \frac{2}{R}\] \[=-\frac{1}{2R}\] As\[f\]is negative, so air bubble behaves as concave lens.You need to login to perform this action.
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