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(D) The focal length of a lens in a medium is given by the formula: $\frac{1}{f_m} = \left( \frac{\mu_l}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \fr

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$65$

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(D) The focal length of a lens in a medium is given by the formula: $\frac{1}{f_m} = \left( \frac{\mu_l}{\mu_m} - 1 ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.Since the focal length in air $(f_a = 20 \ cm)$ is $\frac{1}{f_a} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$,we can write the focal length in a medium as $f_m = f_a \frac{(\mu - 1)}{(\frac{\mu}{\mu_m} - 1)}$.For the upper half,$\mu_m = \mu_1 = 1.2$:$f_1 = 20 	imes \frac{1.5 - 1}{\frac{1.5}{1.2} - 1} = 20 	imes \frac{0.5}{1.25 - 1} = 20 	imes \frac{0.5}{0.25} = 40 \ cm$.For the lower half,$\mu_m = \mu_2 = 2.5$:$f_2 = 20 	imes \frac{1.5 - 1}{\frac{1.5}{2.5} - 1} = 20 	imes \frac{0.5}{0.6 - 1} = 20 	imes \frac{0.5}{-0.4} = -25 \ cm$.Since the object is at infinity,the images are formed at the respective focal points.The separation between the two images is $|f_1| + |f_2| = |40| + |-25| = 65 \ cm$.

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