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The radii of curvature of a lens are +20cm and +30cm. The material of the lens has a refracting index 1.6. Find the focal length of the lens.
  1. If it is placed in air.
  2. If it is placed in water $(\mu=1.33).$
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$\text{R}_1=+20\text{cm}; \ \text{R}_2=+30\text{cm}; \ \mu=1.6$
  1. If placed in air:
$\frac{1}{\text{f}}=(\mu_{\text{g}}-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)=\Big(\frac{1.6}{1}-1\Big)\Big(\frac{1}{20}-\frac{1}{30}\Big)$

$\Rightarrow\text{f}=\frac{60}{6}=100\text{cm}$
  1. If placed in water:
$\frac{1}{\text{f}}=(\mu_{\text{w}}-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$

$\Rightarrow \Big(\frac{1.6}{1.33}-1\Big)\Big(\frac{1}{20}-\frac{1}{30}\Big)$

$\Rightarrow \Big(\frac{1.60}{1.33}-1\Big)\Big[\frac{1}{6}\Big]$

$=\frac{28}{133\times60}\simeq\frac{1}{300}$

$\Rightarrow\text{f}=300\text{cm}$
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