Two similar thin equi-convex lenses, of focal length f each, are kept coaxially in contact with each other such that the local length of the combination is $F_{1} .$ When the space between the two lenses is filled with glycerin (which has the same refractive index $(μ = 1.5)$ as that of glass) then the equivalent focal length is $F_{2} .$ The ratio $F_{1} : F_{2}$ will be

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2 : 1

1 : 2

2 : 3

3 : 4

answer is B.

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Detailed Solution

Two equi convex lenses

$\frac{1}{F} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$ $⇒ F_{1} = \frac{f}{2}$

Two equi convex, one biconcave of same radius of curvature.

$\frac{1}{F_{2}} = \frac{1}{f} - \frac{1}{f} + \frac{1}{f} = \frac{1}{f}$ $⇒ F_{2} = f$

$∴ \frac{F_{1}}{F_{2}} = \frac{1}{2}$

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