There are two thin symmetrical lenses, one is converging with refractive index 2 and the other is diverging with refractive index 1.5. Both lenses have the same radius of curvature of 10 cm. The lenses were put close together and submerged into water. What is the focal length of the system in water?. The refractive index of water is $\frac{4}{3}$

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40 cm

$\frac{20}{3}$ cm

$\frac{40}{3}$ cm

$\frac{- 40}{3}$ cm

answer is B.

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Detailed Solution

The focal length of converging in water is $f_{1}$ .
$∴ \frac{1}{f_{1}} = (\frac{μ_{i}}{μ_{w}} - 1) (\frac{1}{r_{1}} - \frac{1}{r_{2}}) = (\frac{2}{4 / 3} - 1) (\frac{2}{10})$
$= (\frac{6 - 4}{4}) \times \frac{2}{10} = \frac{1}{2} \times \frac{2}{10}$
$∴ f_{1} = 10 c m$
And The focal length of diverging in water is $f_{2}$

$\frac{1}{f_{2}} = (\frac{1.5}{4 / 3} - 1) (\frac{- 2}{10})$
Or $\frac{1}{f_{2}} = \frac{0.5}{4} (\frac{- 2}{10}) = \frac{- 1}{40}$
$∴ f_{2} = - 40 c m$

Combination of lens
$∴ \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1}{10} - \frac{1}{40} = \frac{4 - 1}{40}$
$∴ f = \frac{40}{3} c m$

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