An equiconvex lens of radius of curvature 14 | vedclass

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(D) From the diagram,the lens can be considered as a combination of two plano-convex lenses.For the left half,the focal length $f_1$ is given by:$\fra

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) From the diagram,the lens can be considered as a combination of two plano-convex lenses.For the left half,the focal length $f_1$ is given by:$\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{0.5}{R}$For the right half,the focal length $f_2$ is given by:$\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (1.2 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) = \frac{0.2}{R}$The combined power of the lens is $P = P_1 + P_2$,so $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$.$\frac{1}{f} = \frac{0.5}{R} + \frac{0.2}{R} = \frac{0.7}{14} = \frac{0.7}{14} = 0.05 \ cm^{-1}$.Thus,$f = \frac{1}{0.05} = 20 \ cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $u = -40 \ cm$ and $f = 20 \ cm$:$\frac{1}{20} = \frac{1}{v} - \frac{1}{-40}$$\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{2-1}{40} = \frac{1}{40}$Therefore,$v = 40 \ cm$.

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