A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices μ 1 and μ 2 and R is th

Question

A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces 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 the combination is

TestHub · Accepted Answer

Let the focal length of the plano-convex lens be $f_{1}$ and its refractive index be $μ_{1}$ .

Applying the lens maker's formula for the plano-convex lens, assuming light is incident on the plane surface:
$\frac{1}{f _{1}} = (μ_{1} - 1) (\frac{1}{\infty} - \frac{1}{- R}) = \frac{μ _{1} - 1}{R}$

Let the focal length of the plano-concave lens be $f_{2}$ and its refractive index be $μ_{2}$ . Since the plano-convex lens fits exactly into it, the concave surface of the second lens has the same radius of curvature $R$ and matches the convex surface of the first lens.

Applying the lens maker's formula for the plano-concave lens:
$\frac{1}{f _{2}} = (μ_{2} - 1) (\frac{1}{- R} - \frac{1}{\infty}) = - \frac{( μ _{2} - 1 )}{R}$

The equivalent focal length $F$ of the combination of thin lenses in contact is given by:
$\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$

Substituting the values of $\frac{1}{f _{1}}$ and $\frac{1}{f _{2}}$ :
$\frac{1}{F} = \frac{μ _{1} - 1}{R} - \frac{μ _{2} - 1}{R}$

$\frac{1}{F} = \frac{μ _{1} - 1 - μ _{2} + 1}{R}$

$\frac{1}{F} = \frac{μ _{1} - μ _{2}}{R}$

$F = \frac{R}{μ _{1} - μ _{2}}$ .

Answer: $\frac{R}{( μ _{1} - μ _{2} )}$

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