A glass convex lens (mu_{g}=1.5) has a focal | vedclass

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(D) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For air: $\frac{1}{f_a} = (\mu_g - 1)

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(D) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For air: $\frac{1}{f_a} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = (1.5 - 1) K = 0.5 K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.Given $f_a = 8 \ cm$,so $\frac{1}{8} = 0.5 K \Rightarrow K = \frac{1}{4}$.For water: $\frac{1}{f_w} = (\frac{\mu_g}{\mu_w} - 1) K$.Substituting the values: $\frac{1}{f_w} = (\frac{1.5}{4/3} - 1) 	imes \frac{1}{4} = (\frac{4.5}{4} - 1) 	imes \frac{1}{4} = (1.125 - 1) 	imes \frac{1}{4} = 0.125 	imes \frac{1}{4} = \frac{1}{8} 	imes \frac{1}{4} = \frac{1}{32}$.Therefore,$f_w = 32 \ cm$.

$A$ glass convex lens $(\mu_{g}=1.5)$ has a focal length of $8 \ cm$ when placed in air. What would be the focal length of the lens in $cm$ when it is immersed in water $(\mu_{w}=\frac{4}{3})$?

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