A thin lens made of glass of refractive index | vedclass

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(C) The power of a lens in air is given by the sum of the powers of its surfaces: $P_a = P_1 + P_2 = 11 D - 6 D = 5 D$.Using the lens maker's formula,

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$-0.625 D$

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(C) The power of a lens in air is given by the sum of the powers of its surfaces: $P_a = P_1 + P_2 = 11 D - 6 D = 5 D$.Using the lens maker's formula,the power of a lens in a medium is related to its power in air by the ratio: $\frac{P_l}{P_a} = \frac{_l\mu_g - 1}{_a\mu_g - 1}$.Here,$_a\mu_g = 1.5$ and $_l\mu_g = \frac{_a\mu_g}{_a\mu_l} = \frac{1.5}{1.6} = 0.9375$.Substituting the values: $\frac{P_l}{5} = \frac{0.9375 - 1}{1.5 - 1} = \frac{-0.0625}{0.5} = -0.125$.Therefore,$P_l = 5 	imes (-0.125) = -0.625 D$.

$A$ thin lens made of glass of refractive index $1.5$ has a front surface $+ 11 D$ power and back surface $-6 D.$ If this lens is submerged in a liquid of refractive index $1.6,$ the resulting power of the lens is

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