Q: A Plano convex lens fits exactly into a Plano concave lens. Their plane surface are parallel to each other. If lenses are made of different materials of refractive indices μ_{1} and μ_{2} and R is the radius of curvature of the curved surface of the lenses, then the focal length of combination is

$\displaystyle (a) \frac{2 R}{\mu_2 – \mu_1} $

$\displaystyle (b) \frac{R}{2 (\mu_2 – \mu_1)} $

$\displaystyle (c) \frac{R}{2 (\mu_1 – \mu_2)} $

$\displaystyle (d) \frac{R}{\mu_1 – \mu_2} $

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Sol: Using Lens Maker’s formula ,

$\displaystyle \frac{1}{f_1} = (\mu_1 – 1)( \frac{1}{\infty} – \frac{1}{-R} )$

$\displaystyle \frac{1}{f_1} = \frac{\mu_1 – 1}{R} $

Similarly ,

$\displaystyle \frac{1}{f_2} = (\mu_2 – 1)( \frac{1}{-R} – \frac{1}{\infty} )$

$\displaystyle \frac{1}{f_2} = – \frac{\mu_2 – 1}{R} $

The equivalent focal length (f) of the given lens is

$\displaystyle \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} $

$\displaystyle \frac{1}{f} = \frac{\mu_1 – 1}{R} – \frac{\mu_2 – 1}{R} $

$\displaystyle \frac{1}{f} = \frac{\mu_1 – \mu_2}{R} $

$\displaystyle f = \frac{R}{\mu_1 – \mu_2} $