Two similar thin equi-convex lenses,each of f | vedclass

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(B) For two thin lenses in contact,the equivalent focal length $F_{1}$ is given by $\frac{1}{F_{1}} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$,which i

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$1:2$

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(B) For two thin lenses in contact,the equivalent focal length $F_{1}$ is given by $\frac{1}{F_{1}} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$,which implies $F_{1} = \frac{f}{2}$.When the space between the lenses is filled with a liquid of the same refractive index as the glass,the combination acts as a single block of glass. Since the outer surfaces are convex and the inner space is filled,the system effectively becomes a single convex lens with the same radii of curvature as the original lenses. However,the refractive index of the liquid is $\mu = 1.5$,which is equal to the glass. Thus,the entire system behaves as a single lens with focal length $F_{2} = f$.Therefore,the ratio $\frac{F_{1}}{F_{2}} = \frac{f/2}{f} = \frac{1}{2}$.

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