Two thin lenses of focal lengths f_1 and f_2 | vedclass

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Question

(D) When two thin lenses of focal lengths $f_1$ and $f_2$ are placed in contact,the equivalent focal length $F$ is given by the formula: $\frac{1}{F}

Vedclass · Accepted Answer

$\frac{f_1+f_2}{f_1 f_2}$

Vedclass · Answer

(D) When two thin lenses of focal lengths $f_1$ and $f_2$ are placed in contact,the equivalent focal length $F$ is given by the formula: $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$ The power $P$ of a lens is defined as the reciprocal of its focal length in meters,i.e.,$P = \frac{1}{f}$. The power of the combination $P$ is the sum of the individual powers $P_1$ and $P_2$: $P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2}$ Simplifying the expression: $P = \frac{f_1 + f_2}{f_1 f_2}$ Thus,the correct option is $D$.

Two thin lenses of focal lengths $f_1$ and $f_2$ are in contact and coaxial. The power of the combination is

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