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(A) The system consists of two lenses in contact: a plano-convex glass lens and a plano-concave water lens.$1$. For the glass lens (refractive index $

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$40 \ cm$

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(A) The system consists of two lenses in contact: a plano-convex glass lens and a plano-concave water lens.$1$. For the glass lens (refractive index $n_g = 1.5$) in water (refractive index $n_w = 4/3$):Using the lens maker's formula $\frac{1}{f_1} = (\frac{n_g}{n_w} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$:Here,$R_1 = \infty$ and $R_2 = -15 \ cm$.$\frac{1}{f_1} = (\frac{1.5}{4/3} - 1)(\frac{1}{\infty} - \frac{1}{-15}) = (\frac{4.5}{4} - 1)(\frac{1}{15}) = (1.125 - 1)(\frac{1}{15}) = 0.125 	imes \frac{1}{15} = \frac{1}{8} 	imes \frac{1}{15} = \frac{1}{120} \ cm^{-1}$.$2$. For the water lens (refractive index $n_w = 4/3$) in water (this is effectively a concave surface formed by the air-water interface):Using $\frac{1}{f_2} = (\frac{n_w}{n_w} - 1)...$ is incorrect here. The interface is between air $(n_a = 1)$ and water $(n_w = 4/3)$.$\frac{1}{f_2} = (\frac{n_a}{n_w} - 1)(\frac{1}{R_1} - \frac{1}{R_2}) = (\frac{1}{4/3} - 1)(\frac{1}{-15} - \frac{1}{\infty}) = (0.75 - 1)(-\frac{1}{15}) = (-0.25)(-\frac{1}{15}) = \frac{1}{4} 	imes \frac{1}{15} = \frac{1}{60} \ cm^{-1}$.$3$. Resultant focal length:$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{120} + \frac{1}{60} = \frac{1+2}{120} = \frac{3}{120} = \frac{1}{40} \ cm^{-1}$.Therefore,$f_{eq} = 40 \ cm$.

Find the resultant focal length for the following system where the common radius of curvature is $15 \ cm$. The glass has a refractive index of $1.5$ and water has a refractive index of $\frac{4}{3}$.

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