The focal length of a converging lens in air | vedclass

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(C) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.In air,the focal length $f_a = R$,so $

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$4R$

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(C) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.In air,the focal length $f_a = R$,so $\frac{1}{R} = (\mu_g - 1) K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.Given $\mu_g = 1.5$,we have $\frac{1}{R} = (1.5 - 1) K = 0.5 K$,which implies $K = \frac{2}{R}$.When the lens is dipped in water $(\mu_w = 1.33 \approx 4/3)$,the new focal length $f_w$ is given by $\frac{1}{f_w} = (\frac{\mu_g}{\mu_w} - 1) K$.Substituting the values: $\frac{1}{f_w} = (\frac{1.5}{4/3} - 1) 	imes \frac{2}{R} = (\frac{4.5}{4} - 1) 	imes \frac{2}{R} = (1.125 - 1) 	imes \frac{2}{R} = 0.125 	imes \frac{2}{R} = \frac{0.25}{R} = \frac{1}{4R}$.Therefore,$f_w = 4R$.

The focal length of a converging lens in air is $R$. If it is dipped in water of refractive index $1.33$,then its focal length will be around (Refractive index of lens material is $1.5$).

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