Two identical thin biconvex lenses of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is_______ cm

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Detailed Solution

By lens maker formula, $\frac{1}{f} = (^{g} μ_{a} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

For Lens 1:

$\frac{1}{f_{1}} = (1.5 - 1) (\frac{1}{R} - \frac{1}{- R}) \Rightarrow \frac{1}{f_{1}} = 0.5 \times \frac{2}{R} \Rightarrow f_{1} = R \Rightarrow R = 15 c m$

For Lens 2:

$\frac{1}{f_{2}} = (1.25 - 1) (\frac{1}{- R} - \frac{1}{+ R}) = - 0.25 \times \frac{2}{R} = - \frac{0.5}{R} = \frac{- 1}{2 R}, ∴ f_{2} = - 2 R$

For Lens 3:

Similarly like lens $1, f_{3} = R$ So, $\frac{1}{f_{n e t}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}} = \frac{1}{R} + \frac{1}{- 2 R} + \frac{1}{R}$