Q: Two thin symmetrical lenses of different nature and of different material have equal radii of curvature

R = 15 cm. The lenses are put close together and immersed in water (μ_{w} = 4/3). The focal length of the system in water is 30 cm. The difference between refractive indices of the two lenses is

Sol: Let f_{1} and f_{2} be the focal lengths in water. Then

$\large \frac{1}{f_1} = (\frac{\mu_1}{\mu_w} – 1)(\frac{1}{R} – \frac{1}{R} ) $

$\large \frac{1}{f_1} = (\frac{\mu_1}{\mu_w} – 1)(\frac{2}{R} ) $ ….(i)

$\large \frac{1}{f_2} = (\frac{\mu_2}{\mu_w} – 1)(-\frac{1}{R} – \frac{1}{R} ) $

$\large \frac{1}{f_2} = (\frac{\mu_2}{\mu_w} – 1)(-\frac{2}{R} ) $ …(ii)

Adding Eqs. (i) and (ii) we get

$\large \frac{1}{f_1} + \frac{1}{f_2} = \frac{2(\mu_1 -\mu_2)}{\mu_w R}$

But the given system is equal to combination of two lens kept in contact in liquid so

$\large \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} $

$\large (\mu_1 -\mu_2) = \frac{\mu_w R}{60} $

substituting the values we get ,

$\large (\mu_1 -\mu_2) = \frac{4 \times 15 }{3\times 60} = \frac{1}{3} $