The radius of curvature of a lens is 20,cm an | vedclass

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(D) The system consists of three lenses in contact: two plano-concave liquid lenses and one biconvex glass lens.$1$. For the first liquid lens $(f_1)$

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$-100$

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(D) The system consists of three lenses in contact: two plano-concave liquid lenses and one biconvex glass lens.$1$. For the first liquid lens $(f_1)$: The refractive index of the liquid is $\mu_l = 1.6$. The surfaces are planar and concave with radius $R = 20\,cm$. Using the lens maker's formula: $\frac{1}{f_1} = (\mu_l - 1) (\frac{1}{\infty} - \frac{1}{20}) = (1.6 - 1) (0 - 0.05) = 0.6 	imes (-0.05) = -0.03 = -\frac{3}{100}$.$2$. For the biconvex glass lens $(f_2)$: The refractive index is $\mu_g = 1.5$. The surfaces are convex with $R_1 = 20\,cm$ and $R_2 = -20\,cm$. Using the lens maker's formula: $\frac{1}{f_2} = (\mu_g - 1) (\frac{1}{20} - \frac{1}{-20}) = (1.5 - 1) (\frac{1}{20} + \frac{1}{20}) = 0.5 	imes \frac{2}{20} = \frac{1}{20}$.$3$. For the second liquid lens $(f_3)$: Similar to the first,$\frac{1}{f_3} = (1.6 - 1) (\frac{1}{-20} - \frac{1}{\infty}) = 0.6 	imes (-0.05) = -\frac{3}{100}$.$4$. The equivalent focal length $F$ of the system is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$.Substituting the values: $\frac{1}{F} = -\frac{3}{100} + \frac{1}{20} - \frac{3}{100} = -\frac{6}{100} + \frac{5}{100} = -\frac{1}{100}$.Therefore,$F = -100\,cm$.

The radius of curvature of a lens is $20\,cm$ and its refractive index is $1.5$. If the refractive index of the surrounding liquid is $1.6$,then the focal length of the system is .......... $cm$.

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