A lens which has a focal length of 4 ; cm and | vedclass

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(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a lens in air ($f_a = 4 \; cm$,$\m

Vedclass · Accepted Answer

$-12.8$

Vedclass · Answer

(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a lens in air ($f_a = 4 \; cm$,$\mu_g = 1.4$): $\frac{1}{4} = (1.4 - 1) K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.So,$K = \frac{0.4}{4} = 0.1$.When immersed in a liquid of refractive index $\mu_l = 1.6$,the new focal length $f_l$ is given by $\frac{1}{f_l} = (\frac{\mu_g}{\mu_l} - 1) K$.Substituting the values: $\frac{1}{f_l} = (\frac{1.4}{1.6} - 1) 	imes 0.1$.$\frac{1}{f_l} = (0.875 - 1) 	imes 0.1 = -0.125 	imes 0.1 = -0.0125$.$f_l = -\frac{1}{0.0125} = -80$ (Wait,re-calculating: $\frac{1.4}{1.6} = 0.875$. $0.875 - 1 = -0.125$. $\frac{1}{f_l} = -0.125 	imes 0.1 = -0.0125$. $f_l = -80$).Re-evaluating the provided solution logic: $\frac{f_l}{f_a} = \frac{(\mu_g - 1)}{(\frac{\mu_g}{\mu_l} - 1)}$.$\frac{f_l}{4} = \frac{0.4}{\frac{1.4}{1.6} - 1} = \frac{0.4}{0.875 - 1} = \frac{0.4}{-0.125} = -3.2$.$f_l = 4 	imes (-3.2) = -12.8 \; cm$.

$A$ lens which has a focal length of $4 \; cm$ and a refractive index of $1.4$ is immersed in a liquid of refractive index $1.6$. What will be the new focal length in $cm$?

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