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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
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Verified Answer
The correct answer is:
$\frac{R}{\left(\mu_1-\mu_2\right)}$
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
$$
\begin{aligned}
\frac{1}{f} & =\frac{\left(\mu_1-1\right)}{R}-\frac{\left(\mu_2-1\right)}{R} \\
& =\frac{\left[\mu_1-1-\mu_2+1\right]}{R}=\frac{\mu_1-\mu_2}{R}
\end{aligned}
$$
$$
\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}
$$
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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