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

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(B) Using the lens maker's formula $\frac{1}{f} = (n-1)(\frac{1}{R_1} - \frac{1}{R_2})$:For the first plano-convex lens: $\frac{1}{f_1} = (1.5 - 1)(\f

Vedclass · Accepted Answer

$40$

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(B) Using the lens maker's formula $\frac{1}{f} = (n-1)(\frac{1}{R_1} - \frac{1}{R_2})$:For the first plano-convex lens: $\frac{1}{f_1} = (1.5 - 1)(\frac{1}{14} - \frac{1}{\infty}) = \frac{0.5}{14} = \frac{1}{28} \, cm^{-1}$.For the second plano-convex lens: $\frac{1}{f_2} = (1.2 - 1)(\frac{1}{\infty} - \frac{1}{-14}) = \frac{0.2}{14} = \frac{1}{70} \, cm^{-1}$.The equivalent focal length $f_{eq}$ is given by $\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{28} + \frac{1}{70} = \frac{5 + 2}{140} = \frac{7}{140} = \frac{1}{20} \, cm^{-1}$.Thus,$f_{eq} = 20 \, cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with $u = -40 \, cm$ and $f = 20 \, cm$:$\frac{1}{v} - \frac{1}{-40} = \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{2-1}{40} = \frac{1}{40}$.Therefore,$v = 40 \, cm$.

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