Two similar thin equi-convex lenses, of focal length $f$ each, are kept coaxially in contact with each other s

Question

Two similar thin equi-convex lenses, of focal length $f$ each, are kept coaxially in contact with each other such that the focal length of the combination is $F_1$. When the space between the two lenses is filled with glycerin which has the same refractive index as that of glass ($\mu = 1.5$), then the equivalent focal length is $F_2$. The ratio $F_1 : F_2$ will be:

Accepted Answer

1. **Initial Combination ($F_1$):** 
   - For an equi-convex lens with refractive index $\mu=1.5$ and radius of curvature $R$, the focal length $f$ is given by Lens Maker's Formula: $\frac{1}{f} = (\mu-1)(\frac{2}{R}) = (1.5-1)\frac{2}{R} = \frac{1}{R}$. Thus, $f = R$.
   - When two such lenses are in contact, the equivalent focal length $F_1$ is: $\frac{1}{F_1} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$. Therefore, $F_1 = \frac{f}{2}$.
2. **With Glycerin ($F_2$):** 
   - The space between the two convex lenses forms an equi-concave liquid lens. Since the refractive index of glycerin ($1.5$) is the same as the glass, the liquid lens behaves as a concave lens with the same radius of curvature $R$.
   - Focal length of the liquid lens ($f_l$): $\frac{1}{f_l} = (1.5-1)(\frac{-1}{R} - \frac{1}{R}) = 0.5(\frac{-2}{R}) = -\frac{1}{R} = -\frac{1}{f}$. So, $f_l = -f$.
   - The new system consists of three lenses in contact: Convex ($f$) + Concave ($f_l = -f$) + Convex ($f$).
   - The equivalent focal length $F_2$ is: $\frac{1}{F_2} = \frac{1}{f} + \frac{1}{-f} + \frac{1}{f} = \frac{1}{f}$. Therefore, $F_2 = f$.
3. **Calculate Ratio:** 
   - Ratio $F_1 : F_2 = \frac{f}{2} : f = 1 : 2$.

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