The refractive index of a glass convex lens i | vedclass

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(B) Given: Refractive index of glass $\mu_g = 1.5$,refractive index of air $\mu_a = 1$,refractive index of liquid $\mu_l = 1.25$,and radii of curvatur

Vedclass · Accepted Answer

$2:5$

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(B) Given: Refractive index of glass $\mu_g = 1.5$,refractive index of air $\mu_a = 1$,refractive index of liquid $\mu_l = 1.25$,and radii of curvature $R_1 = 20 ~cm, R_2 = -20 ~cm$.Using the Lens Maker's Formula: $P = \frac{1}{f} = (\frac{\mu_g}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.Power in air $(P_a)$: $P_a = (\frac{1.5}{1} - 1)(\frac{1}{20} - \frac{1}{-20}) = (0.5)(\frac{2}{20}) = 0.5 	imes 0.1 = 0.05 ~cm^{-1}$.Power in liquid $(P_l)$: $P_l = (\frac{1.5}{1.25} - 1)(\frac{1}{20} - \frac{1}{-20}) = (1.2 - 1)(0.1) = 0.2 	imes 0.1 = 0.02 ~cm^{-1}$.Ratio of power in liquid to power in air: $\frac{P_l}{P_a} = \frac{0.02}{0.05} = \frac{2}{5}$.

The refractive index of a glass convex lens is $1.5$. The radius of curvature of each of the two surfaces of the lens is $20 ~cm$. The ratio of the power of the lens when immersed in a liquid of refractive index $1.25$ to that when placed in air is:

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