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

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(B) The lens maker's formula is given by $\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.For the first plano-convex lens:$n_1 = 1.5$,$

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) The lens maker's formula is given by $\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.For the first plano-convex lens:$n_1 = 1.5$,$R_1 = 14\; cm$,$R_2 = \infty$.$\frac{1}{f_1} = (1.5 - 1)\left(\frac{1}{14} - \frac{1}{\infty}\right) = \frac{0.5}{14}$.For the second plano-convex lens:$n_2 = 1.2$,$R_1 = \infty$,$R_2 = -14\; cm$.$\frac{1}{f_2} = (1.2 - 1)\left(\frac{1}{\infty} - \frac{1}{-14}\right) = \frac{0.2}{14}$.The focal length $f$ of the combination is given by $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$.$\frac{1}{f} = \frac{0.5}{14} + \frac{0.2}{14} = \frac{0.7}{14} = \frac{1}{20}$.Thus,$f = 20\; cm$.Using the thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:Given $u = -40\; cm$ and $f = 20\; cm$.$\frac{1}{v} - \frac{1}{-40} = \frac{1}{20}$.$\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{2-1}{40} = \frac{1}{40}$.Therefore,$v = 40\; cm$.

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