A glass convex lens (mu_g = 1.5) has a focal | vedclass

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(D) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.In air $(f_a = 8\, cm)$: $\frac{1}{8} = (1.

Vedclass · Accepted Answer

$32$

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(D) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.In air $(f_a = 8\, cm)$: $\frac{1}{8} = (1.5 - 1) K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.So,$K = \frac{1}{8 	imes 0.5} = \frac{1}{4}$.In water,the refractive index of the lens relative to water is $\mu_{rel} = \frac{\mu_g}{\mu_w} = \frac{1.5}{1.33} \approx 1.128$.The new focal length $f_w$ is given by: $\frac{1}{f_w} = (\frac{\mu_g}{\mu_w} - 1) K$.$\frac{1}{f_w} = (\frac{1.5}{1.33} - 1) 	imes \frac{1}{4} = (\frac{1.5 - 1.33}{1.33}) 	imes \frac{1}{4} = \frac{0.17}{1.33 	imes 4} = \frac{0.17}{5.32} \approx \frac{1}{31.29}$.Using the standard approximation $\mu_w = 4/3$,we get: $\frac{1}{f_w} = (\frac{1.5}{4/3} - 1) 	imes \frac{1}{4} = (1.125 - 1) 	imes \frac{1}{4} = 0.125 	imes \frac{1}{4} = \frac{1}{8} 	imes \frac{1}{4} = \frac{1}{32}$.Therefore,$f_w = 32\, cm$.

$A$ glass convex lens $(\mu_g = 1.5)$ has a focal length of $8\, cm$ in air. What will be its focal length when it is immersed in water (in $, cm$)? $(\mu_w = 1.33)$

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