Q127 / 180 · NEET 2013 (All India) · Physics

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

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

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

Hence, the correct answer is option: 'C'.

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