The two surfaces of a biconvex lens have the | vedclass

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(D) For a biconvex lens of focal length $f$ and refractive index $\mu_g = 1.5$,the lens maker's formula in air is $\frac{1}{f} = (\mu_g - 1)(\frac{1}{

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$40$

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(D) For a biconvex lens of focal length $f$ and refractive index $\mu_g = 1.5$,the lens maker's formula in air is $\frac{1}{f} = (\mu_g - 1)(\frac{1}{R} - \frac{1}{-R}) = (0.5)(\frac{2}{R}) = \frac{1}{R}$. Given $f = 10 ~cm$,we have $R = 10 ~cm$.When the lens is cut perpendicular to the principal axis,each plano-convex lens has a focal length $f'$ given by $\frac{1}{f'} = (\mu_g - 1)(\frac{1}{R} - \frac{1}{\infty}) = \frac{0.5}{R} = \frac{0.5}{10} = \frac{1}{20}$. Thus,$f' = 20 ~cm$.When these two plano-convex lenses are joined such that their convex surfaces touch,the combination acts as a biconvex lens with the same focal length $F = 10 ~cm$ in air.When immersed in water (refractive index $\mu_w = 4/3$),the new focal length $F'$ is given by $\frac{1}{F'} = (\frac{\mu_g}{\mu_w} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.For the combination,the effective power is $\frac{1}{F'} = (\frac{1.5}{4/3} - 1)(\frac{1}{R} - \frac{1}{-R}) = (\frac{4.5}{4} - 1)(\frac{2}{R}) = (1.125 - 1)(\frac{2}{10}) = 0.125 	imes 0.2 = 0.025$.Therefore,$F' = \frac{1}{0.025} = 40 ~cm$.

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