A bi-convex lens is formed with two thin plan | vedclass

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(B) The effective focal length $f_{eff}$ of the combination of two thin lenses in contact is given by $\frac{1}{f_{eff}} = \frac{1}{f_1} + \frac{1}{f_

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) The effective focal length $f_{eff}$ of the combination of two thin lenses in contact is given by $\frac{1}{f_{eff}} = \frac{1}{f_1} + \frac{1}{f_2}$.For a plano-convex lens,the lens maker's formula is $\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For the first lens,$n_1 = 1.5$,$R_1 = 14\, cm$,and $R_2 = \infty$,so $\frac{1}{f_1} = (1.5 - 1) \left( \frac{1}{14} - 0 \right) = \frac{0.5}{14} = \frac{1}{28}$.For the second lens,$n_2 = 1.2$,$R_1 = \infty$,and $R_2 = -14\, cm$,so $\frac{1}{f_2} = (1.2 - 1) \left( 0 - \frac{1}{-14} \right) = \frac{0.2}{14} = \frac{1}{70}$.Thus,$\frac{1}{f_{eff}} = \frac{1}{28} + \frac{1}{70} = \frac{5 + 2}{140} = \frac{7}{140} = \frac{1}{20}$.So,$f_{eff} = 20\, cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where $u = -40\, cm$ and $f = 20\, cm$:$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} - \frac{1}{40} = \frac{2 - 1}{40} = \frac{1}{40}$.Therefore,$v = 40\, cm$.

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