Four combinations of two thin lenses are given in List-I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in List-I with their focal length in List-II and select the correct answer using the code given below the lists.

List– I

List– II
A
I
2r
B
II
r/2
C
III
-r
D
IV
r
Codes:
A B C D
1 I II III IV
2 II IV III I
3 IV I II III
4 II I III IV

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answer is B.

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Detailed Solution

For double convex lens, $(P) \frac{1}{f} = (g m - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$
$\begin{array}{l} \Rightarrow (1.5 - 1) (\frac{1}{r} - \frac{1}{r}) = (1.5 - 1) [\frac{2}{r}] = \frac{1}{r} \Rightarrow f = r \\ \frac{1}{F_{e q}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r} \\ ∴ F_{e q} = \frac{r}{2} \end{array}$
For (Q) plano-convex lens $\frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$
$\begin{array}{l} = (1.5 - 1) [\frac{1}{\infty} - \frac{1}{- r}] = \frac{0.5}{r} = \frac{1}{2 r} ∴ f = 2 r \\ \frac{1}{F_{e q}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1}{2 r} + \frac{1}{2 r} = \frac{2}{2 r} = \frac{1}{r} ∴ F_{e q} = r \end{array}$
For (R) plano-concave lens
$\begin{array}{l} \frac{1}{f} = (1.5 - 1) (\frac{1}{- r} - \frac{1}{\infty}) \Rightarrow f = - 2 r \\ \frac{1}{F_{e q}} = \frac{1}{f} + \frac{1}{f} = \frac{1}{- 2 r} + \frac{1}{- 2 r} \Rightarrow F_{e q} = - r \end{array}$