A convex lens of refractive index 1.5 and foc | vedclass

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(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.In air: $\frac{1}{f_{	ext{air}}} = (1.5 -

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$3$

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(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.In air: $\frac{1}{f_{	ext{air}}} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = 0.5 	imes K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.Given $f_{	ext{air}} = f = 18 \ cm$,so $K = \frac{1}{0.5f} = \frac{2}{f}$.In water $(\mu_w = 4/3)$: $\frac{1}{f_{	ext{water}}} = (\frac{1.5}{4/3} - 1) K = (\frac{4.5}{4} - 1) K = (1.125 - 1) K = 0.125 K$.Substituting $K = \frac{2}{f}$: $\frac{1}{f_{	ext{water}}} = 0.125 	imes \frac{2}{f} = \frac{0.25}{f} = \frac{1}{4f}$.Thus,$f_{	ext{water}} = 4f$.The difference in focal lengths is $f_{	ext{water}} - f_{	ext{air}} = 4f - f = 3f$.Comparing with $\alpha 	imes f$,we get $\alpha = 3$.

$A$ convex lens of refractive index $1.5$ and focal length $f = 18 \ cm$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$. The value of $\alpha$ is . . . . . . . (Refractive index of water $= 4/3$)

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