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(B) The lens maker's formula is given by $\frac{1}{f} = (n_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.In air $(n_a = 1)$: $\frac{1}{f_a} = (

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$288$

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(B) The lens maker's formula is given by $\frac{1}{f} = (n_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.In air $(n_a = 1)$: $\frac{1}{f_a} = (n_g - 1) K$, where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.Given $f_a = 12 	ext{ cm}$ and $n_g = 1.6$, we have $\frac{1}{12} = (1.6 - 1) K = 0.6 K$, so $K = \frac{1}{12 	imes 0.6} = \frac{1}{7.2}$.In water $(n_w = 1.28)$: $\frac{1}{f_w} = (\frac{n_g}{n_w} - 1) K$.Substituting the values: $\frac{1}{f_w} = (\frac{1.6}{1.28} - 1) 	imes \frac{1}{7.2} = (1.25 - 1) 	imes \frac{1}{7.2} = 0.25 	imes \frac{1}{7.2} = \frac{1}{4 	imes 7.2} = \frac{1}{28.8}$.Thus, $f_w = 28.8 	ext{ cm} = 288 	ext{ mm}$.

$A$ convex lens having refractive index $1.6$ has a focal length of $12 \text{ cm}$ when it is in air. The focal length of that lens when placed in water is (refractive index of water = $1.28$) (in $\text{ mm}$)

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