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

• A. $\frac{R}{2(μ_1 +μ_2)}$
• B. $\frac{R}{2(μ_1 - μ_2)}$
• C. $\frac{R}{(μ_1 - μ_2)}$
• D. $\frac{2R}{(μ_2 - μ_1)}$

Answer 1

The Correct Option is C

To determine the focal length of the combination of a plano-convex lens and a plano-concave lens, we need to understand how these lenses interact when placed together. The given lenses have their plane surfaces parallel and curved surfaces fitting together.

The formula for the focal length f of a lens made from a material with refractive index μ and radius of curvature R is provided by the lens maker's formula:

f = \frac{R}{(μ - 1)}

For a plano-convex lens with refractive index μ_1:

f_1 = \frac{R}{(μ_1 - 1)}

For a plano-concave lens with refractive index μ_2, the formula is similar, but since it is concave, it will have a negative focal length:

f_2 = - \frac{R}{(μ_2 - 1)}

When these lenses are combined, the effective focal length F of the system is given by the formula for the combination of thin lenses:

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

Substituting the values for f_1 and f_2:

\frac{1}{F} = \frac{1}{\left(\frac{R}{μ_1 - 1}\right)} - \frac{1}{\left(\frac{R}{μ_2 - 1}\right)}

Simplifying the expression:

\frac{1}{F} = \frac{μ_1 - 1}{R} - \frac{μ_2 - 1}{R}

\frac{1}{F} = \frac{(μ_1 - μ_2)}{R}

Thus, the focal length of the system is:

F = \frac{R}{(μ_1 - μ_2)}

Therefore, the correct answer is: \frac{R}{(μ_1 - μ_2)}, which matches the given correct option.