Two thin lenses of focal length f_{1} and f_{ | vedclass

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(D) When two thin lenses of focal lengths $f_{1}$ and $f_{2}$ are placed in contact,the equivalent focal length $f$ of the combination is given by the

Vedclass · Accepted Answer

$\frac{f_{1}+f_{2}}{f_{1} f_{2}}$

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(D) When two thin lenses of focal lengths $f_{1}$ and $f_{2}$ are placed in contact,the equivalent focal length $f$ of the combination is given by the formula: $\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$.Since the power of a lens is defined as $P = \frac{1}{f}$ (where $f$ is in meters),the power of the combination $P$ is the sum of the individual powers: $P = P_{1} + P_{2}$.Substituting the expressions for power,we get: $P = \frac{1}{f_{1}} + \frac{1}{f_{2}}$.Taking the common denominator,we obtain: $P = \frac{f_{1} + f_{2}}{f_{1} f_{2}}$.Thus,the correct option is $D$.

Two thin lenses of focal length $f_{1}$ and $f_{2}$ are in contact and coaxial. The power of the combination is . . . . . . .

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