A convex lens made of glass (refractive index | vedclass

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(D) The lens maker's formula is given by $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a convex lens,let $R_1 = R$

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(D) The lens maker's formula is given by $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a convex lens,let $R_1 = R$ and $R_2 = -R$. Thus,$\frac{1}{f} = (\mu - 1) \frac{2}{R}$.In air: $\frac{1}{24} = (1.5 - 1) \frac{2}{R} = 0.5 	imes \frac{2}{R} = \frac{1}{R}$. So,$R = 24 \ cm$.In water: $\frac{1}{f'} = \left( \frac{\mu_g}{\mu_w} - 1 ight) \frac{2}{R} = \left( \frac{1.5}{1.33} - 1 ight) \frac{2}{24}$.Using $\mu_w = 1.33 \approx \frac{4}{3}$,we get $\frac{1}{f'} = \left( \frac{1.5}{4/3} - 1 ight) \frac{1}{12} = \left( \frac{4.5}{4} - 1 ight) \frac{1}{12} = (1.125 - 1) \frac{1}{12} = 0.125 	imes \frac{1}{12} = \frac{1}{8} 	imes \frac{1}{12} = \frac{1}{96}$.Therefore,$f' = 96 \ cm$.

$A$ convex lens made of glass (refractive index $= 1.5$) has a focal length of $24 \ cm$ in air. When it is totally immersed in water (refractive index $= 1.33$),its focal length changes to: (in $cm$)

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