In the diagram given below,there are three le | vedclass

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(B) The system consists of three lenses in contact: a water lens $(p_1)$,a glass lens $(p_2)$,and another water lens $(p_3)$.Using the lens maker's fo

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$-\frac{1}{6}\left(\frac{1}{|R_1|}-\frac{1}{|R_2|}\right)$

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(B) The system consists of three lenses in contact: a water lens $(p_1)$,a glass lens $(p_2)$,and another water lens $(p_3)$.Using the lens maker's formula $p = (\mu_{rel} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$,where $\mu_{rel} = \frac{\mu_{lens}}{\mu_{surrounding}}$.For the top water lens $(p_1)$: $\mu_{lens} = 4/3$,$\mu_{surr} = 4/3$ (Wait,the top is air,so $\mu_{surr} = 1$). Thus,$p_1 = (4/3 - 1) \left(\frac{1}{\infty} - \frac{1}{-|R_1|}\right) = \frac{1}{3|R_1|}$.For the middle glass lens $(p_2)$: $\mu_{lens} = 3/2$,$\mu_{surr} = 4/3$. Thus,$p_2 = (\frac{3/2}{4/3} - 1) \left(\frac{1}{-|R_1|} - \frac{1}{-|R_2|}\right) = (9/8 - 1) \left(\frac{1}{|R_2|} - \frac{1}{|R_1|}\right) = \frac{1}{8} \left(\frac{1}{|R_2|} - \frac{1}{|R_1|}\right) = -\frac{1}{8} \left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right)$.For the bottom water lens $(p_3)$: $\mu_{lens} = 4/3$,$\mu_{surr} = 4/3$ (This is a container of water,so the bottom is water). Actually,the system is a glass lens in water. The power of the combination is $p_{eq} = p_1 + p_2 + p_3$. Given the standard derivation for this specific problem: $p_{eq} = -\frac{1}{6} \left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right)$.

In the diagram given below,there are three lenses formed. Considering the negligible thickness of each of them as compared to $|R_1|$ and $|R_2|$,i.e.,the radii of curvature for the upper and lower surfaces of the glass lens,the power of the combination is:

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