A convex lens of glass (mu_g = 1.45) has a fo | vedclass

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(A) The focal length of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu_{rel} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.In air

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$3.9$

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(A) The focal length of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu_{rel} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}ight)$.In air, the focal length $f_a$ is: $\frac{1}{f_a} = (\mu_g - 1) K$, where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.Given $\mu_g = 1.45$, so $\frac{1}{f_a} = (1.45 - 1) K = 0.45 K$.In the liquid, the focal length $f_l$ is: $\frac{1}{f_l} = (\frac{\mu_g}{\mu_l} - 1) K$.Given $\mu_l = 1.3$, so $\frac{1}{f_l} = (\frac{1.45}{1.3} - 1) K = (\frac{1.45 - 1.3}{1.3}) K = (\frac{0.15}{1.3}) K$.Now, find the ratio $f_l / f_a$:$\frac{f_l}{f_a} = \frac{0.45 K}{(\frac{0.15}{1.3}) K} = \frac{0.45 	imes 1.3}{0.15} = 3 	imes 1.3 = 3.9$.

$A$ convex lens of glass $(\mu_g = 1.45)$ has a focal length $f_a$ in air. The lens is immersed in a liquid of refractive index $(\mu_l) = 1.3$. The ratio of the $f_l / f_a$ is

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