Two identical equiconvex lenses,each of focal | vedclass

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(B) The given combination consists of three lenses in contact: two identical equiconvex glass lenses ($L_1$ and $L_3$) and one water lens $(L_2)$ form

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$\frac{f}{2} < F < f$

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(B) The given combination consists of three lenses in contact: two identical equiconvex glass lenses ($L_1$ and $L_3$) and one water lens $(L_2)$ formed in the space between them.Let $f_1$ be the focal length of each glass lens and $f_2$ be the focal length of the water lens.The combination of lenses is in contact,so the equivalent focal length $F$ is given by:$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$Since $L_1$ and $L_3$ are identical,$f_1 = f_3 = f$. The water lens $L_2$ is a biconcave lens,so its focal length $f_2$ is negative.Thus,$\frac{1}{F} = \frac{1}{f} + \frac{1}{f_2} + \frac{1}{f} = \frac{2}{f} + \frac{1}{f_2}$.Let $R$ be the radius of curvature. For the glass lenses,$\frac{1}{f} = (\mu_g - 1)(\frac{2}{R})$. For the water lens,$\frac{1}{f_2} = -(\mu_w - 1)(\frac{2}{R})$.Since $\mu_g > \mu_w$,the magnitude of the power of the glass lens is greater than that of the water lens,meaning $|\frac{1}{f}| > |\frac{1}{f_2}|$.Therefore,$\frac{1}{F} = \frac{2}{f} - |\frac{1}{f_2}|$. Since $|\frac{1}{f_2}| > 0$,$\frac{1}{F} \frac{f}{2}$.Also,since $\frac{1}{F} = \frac{2}{f} - |\frac{1}{f_2}|$,it is clear that $\frac{1}{F} > \frac{1}{f}$ (because we are subtracting a positive value from $\frac{2}{f}$ but the net power is still positive and less than $\frac{2}{f}$),thus $F Combining these,we get $\frac{f}{2} < F < f$.

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