When a convex lens is immersed in two differe | vedclass

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(C) The lens maker's formula is given by $\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$,where $\mu_l$ is the refractive inde

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$1.65$

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(C) The lens maker's formula is given by $\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$,where $\mu_l$ is the refractive index of the lens and $\mu_m$ is the refractive index of the medium.Let $\mu_l$ be the refractive index of the lens material and $K = (\frac{1}{R_1} - \frac{1}{R_2})$.For the first liquid with refractive index $\mu_{m1} = 1.25$,the focal length $f_1$ is $\frac{1}{f_1} = (\frac{\mu_l}{1.25} - 1)K$.For the second liquid with refractive index $\mu_{m2} = 1.5$,the focal length $f_2$ is $\frac{1}{f_2} = (\frac{\mu_l}{1.5} - 1)K$.The ratio of focal lengths is given as $\frac{f_1}{f_2} = \frac{5}{16}$.Therefore,$\frac{f_1}{f_2} = \frac{(\frac{\mu_l}{1.5} - 1)}{(\frac{\mu_l}{1.25} - 1)} = \frac{5}{16}$.$\frac{\mu_l - 1.5}{1.5} 	imes \frac{1.25}{\mu_l - 1.25} = \frac{5}{16}$.$\frac{\mu_l - 1.5}{\mu_l - 1.25} = \frac{5}{16} 	imes \frac{1.5}{1.25} = \frac{5}{16} 	imes 1.2 = \frac{6}{16} = \frac{3}{8}$.$8(\mu_l - 1.5) = 3(\mu_l - 1.25)$.$8\mu_l - 12 = 3\mu_l - 3.75$.$5\mu_l = 8.25$.$\mu_l = 1.65$.

When a convex lens is immersed in two different liquids of refractive indices $1.25$ and $1.5$,the ratio of the focal lengths of the lens is $5:16$. The refractive index of the material of the lens is

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