A plano-convex lens fits exactly into a plano-concave lens. Their...

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

\mu_1

and

\mu_2

and

R

is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is

Options

A\frac{R}{2 $\left(\mu_1+\mu_2\right)}$
B\frac{R}{2 $\left(\mu_1-\mu_2\right)}$
C\frac{R}{ $\left(\mu_1-\mu_2\right)}$
D\frac{2 R}{ $\left(\mu_2-\mu_1\right)}$

Correct answer

C. \frac{R}{ $\left(\mu_1-\mu_2\right)}$

Step-by-step solution

Focal length of the combination

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

We have

f_1=\frac{R}{\left(\mu_1-1\right)}

and

f_2=\frac{R}{\left(\mu_2-1\right)} or \quad \frac{1}{f_1}=\frac{R}{\left(\mu_1-1\right)} or \frac{1}{f_2}=-\frac{R}{\left(\mu_2-1\right)}

Putting these values in Eq. (i), we get

\begin{aligned}\frac{1}{f}& =\frac{\left(\mu_1-1\right)}{R}-\frac{\left(\mu_2-1\right)}{R}\\ & =\frac{\left[\mu_1-1-\mu_2+1\right]}{R}=\frac{\mu_1-\mu_2}{R} \end{aligned}

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