A lens having refractive index 1.6 has a foca | vedclass

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(B) Using the lens maker's formula: $\frac{1}{f} = (\frac{\mu_L}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.For air,$\mu_m = 1$ and $f = 12 \ cm$:$\f

Vedclass · Accepted Answer

$288$

Vedclass · Answer

(B) Using the lens maker's formula: $\frac{1}{f} = (\frac{\mu_L}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.For air,$\mu_m = 1$ and $f = 12 \ cm$:$\frac{1}{12} = (1.6 - 1)(\frac{1}{R_1} - \frac{1}{R_2})$$\frac{1}{12} = 0.6(\frac{1}{R_1} - \frac{1}{R_2})$$(\frac{1}{R_1} - \frac{1}{R_2}) = \frac{1}{12 	imes 0.6} = \frac{1}{7.2} \ cm^{-1}$.For water,$\mu_m = 1.28$:$\frac{1}{f_w} = (\frac{1.6}{1.28} - 1)(\frac{1}{7.2})$$\frac{1}{f_w} = (1.25 - 1)(\frac{1}{7.2}) = 0.25 	imes \frac{1}{7.2} = \frac{1}{4 	imes 7.2} = \frac{1}{28.8} \ cm^{-1}$.Thus,$f_w = 28.8 \ cm = 288 \ mm$.

$A$ lens having refractive index $1.6$ has a focal length of $12 \ cm$ when it is in air. Find the focal length of the lens when it is placed in water. (Take the refractive index of water as $1.28$) (in $mm$)

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