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Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is
A
-20 cm
B
-25 cm
C
-50 cm
D
50 cm
Detailed Solution
$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$; lens maker's formula $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
$\frac{1}{f_1} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{\infty}\right) = \frac{1}{40}$ $cm^{-1}$
$\frac{1}{f_2} = (1.7 - 1)\left(-\frac{1}{20} - \frac{1}{20}\right) = -\frac{7}{100}$ $cm^{-1}$
$\frac{1}{f_3} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{\infty}\right) = \frac{1}{40}$ $cm^{-1}$
$\frac{1}{f} = \frac{1}{40} - \frac{7}{100} + \frac{1}{40} = \frac{5 - 14 + 5}{200} = -\frac{1}{50}$ $cm^{-1}$
$f = -50$ cm