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) K

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$32$

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,$\frac{1}{f_a} = (\mu_g - 1) K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.Given $f_a = 8\,cm$ and $\mu_g = 1.5$,we have $\frac{1}{8} = (1.5 - 1) K \Rightarrow K = \frac{1}{4}$.When immersed in water,the focal length $f_w$ is given by $\frac{1}{f_w} = (\frac{\mu_g}{\mu_w} - 1) K$.Substituting the values,$\frac{1}{f_w} = (\frac{1.5}{1.33} - 1) 	imes \frac{1}{4}$.Since $1.33 \approx \frac{4}{3}$,we have $\frac{1.5}{1.33} \approx 1.5 	imes \frac{3}{4} = 1.125$.Thus,$\frac{1}{f_w} = (1.125 - 1) 	imes \frac{1}{4} = 0.125 	imes \frac{1}{4} = \frac{1}{8} 	imes \frac{1}{4} = \frac{1}{32}$.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 when it is immersed in water $(\mu_w = 1.33)$ (in $,cm$)?

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