# A Plano convex lens fits exactly into a Plano concave lens. Their plane surface are parallel to each other….

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 μ1 and μ2 and R is the radius of curvature of the curved surface of the lenses, then the focal length of combination is

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

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

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

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

Ans: (d)

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

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

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

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

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