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Q.
An air bubble is contained inside water. It behaves as a
Bihar CECEBihar CECE 2008Ray Optics and Optical Instruments
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 $\frac{1}{f}=\frac{-\frac{1}{3}}{\frac{4}{3}} \times \frac{2}{R}$
$=-\frac{1}{2 R}$
As $f$ is negative, so air bubble behaves as concave lens.
