A convex lens of refractive index 1.5 and foc | vedclass

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(C) 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} = (\mu_g - 1) \l

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(C) 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,$\frac{1}{f_a} = (\mu_g - 1) \left( \frac{2}{R} ight) = (1.5 - 1) \left( \frac{2}{R} ight) = 0.5 \left( \frac{2}{R} ight) = \frac{1}{R}$.Given $f_a = 18 \, cm$,so $\frac{1}{18} = \frac{1}{R}$,which means $R = 18 \, cm$.When immersed in water,$\frac{1}{f_w} = \left( \frac{\mu_g}{\mu_w} - 1 ight) \left( \frac{2}{R} ight)$.Substituting the values: $\frac{1}{f_w} = \left( \frac{1.5}{4/3} - 1 ight) \left( \frac{2}{18} ight) = \left( \frac{4.5}{4} - 1 ight) \left( \frac{1}{9} ight) = \left( 1.125 - 1 ight) \left( \frac{1}{9} ight) = 0.125 	imes \frac{1}{9} = \frac{1}{8} 	imes \frac{1}{9} = \frac{1}{72}$.Thus,$f_w = 72 \, cm$.The change in focal length is $\Delta f = f_w - f_a = 72 - 18 = 54 \, cm$.

$A$ convex lens of refractive index $1.5$ and focal length $18 \, cm$ in air is immersed in water. The change in focal length of the lens will be $........... \, cm$. (Given refractive index of water $= 4/3$)

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