Two identical equiconvex lenses, each of foca | vedclass

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Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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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