A double convex lens of focal length 20 cm i | vedclass

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Question

(A) 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,$\frac{1}{f_a} = (_{a}\mu_{g} -

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) 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,$\frac{1}{f_a} = (_{a}\mu_{g} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.Given $f_a = 20 \ cm$,so $0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = \frac{1}{20}$,which means $\left( \frac{1}{R_1} - \frac{1}{R_2} ight) = \frac{1}{10}$.In water,$\frac{1}{f_w} = (_{w}\mu_{g} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$,where $_{w}\mu_{g} = \frac{_{a}\mu_{g}}{_{a}\mu_{w}} = \frac{1.5}{4/3} = \frac{1.5 	imes 3}{4} = \frac{4.5}{4} = 1.125$.So,$\frac{1}{f_w} = (1.125 - 1) 	imes \frac{1}{10} = 0.125 	imes \frac{1}{10} = \frac{0.125}{10} = \frac{1}{80}$.Therefore,$f_w = 80 \ cm$.

$A$ double convex lens of focal length $20 \ cm$ is made of glass of refractive index $3/2$. When placed completely in water $(_{a}\mu_{w} = 4/3)$,its focal length will be.....$cm$.

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