A transparent thin film of uniform thickness and refractive index $n_{1} = 1.4$ is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index $n_{2} = 1.5$ , as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance $f_{1}$ from the film, while rays of light traversing from glass to air get focused at distance $f_{2}$ from the film, Then

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$| f_{1} | = 3 R$

$| f_{1} | = 2.8 R$

$| f_{2} | = 2 R$

$| f_{2} | = 1.4 R$

answer is A, C.

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$C a s e - I; \frac{n_{1}}{V^{'}} - \frac{1}{\infty} = \frac{n_{1} - 1}{R} \dots . (i) \frac{1.4}{V^{'}} = \frac{1.4 - 1}{R} \Rightarrow \frac{1.4}{V^{'}} = \frac{0.4}{R} V^{'} = \frac{7 R}{2} \frac{n_{2}}{f_{1}} - \frac{n_{1}}{V^{'}} = \frac{n_{2} - n_{1}}{R} \dots (2) \frac{n_{2}}{f_{1}} = \frac{n_{1} - 1}{R} + \frac{n_{2} - n_{1}}{R} \frac{n_{2}}{f_{1}} = \frac{n_{1} - 1 + n_{2} - n_{1}}{R} = \frac{n_{2} - 1}{R} f_{1} = \frac{n_{2} R}{n_{2} - 1} = \frac{1.5 R}{1.5 - 1} = \frac{1.5 R}{0.5} f_{1} = 3 R$

$C a s e - i i : \frac{n_{1}}{V^{'}} - \frac{n_{2}}{\infty} = \frac{n_{1} - n_{2}}{- R} \frac{n_{1}}{V^{'}} = \frac{n_{2} - n_{1}}{R} \dots (1) \frac{1}{f_{2}} - \frac{n_{1}}{V^{'}} = \frac{1 - n_{1}}{- R} = \frac{n_{1} - 1}{R} \dots . (2) (1) + (2) : \frac{1}{f_{2}} = \frac{n_{1} - 1 + n_{2} - n_{1}}{R} = \frac{n_{2} - 1}{R} \Rightarrow f_{2} = 2 R f_{2} = \frac{R}{n_{2} - 1} = \frac{R}{0.5} = 2 R$