Two lens of focal length $f_{1}$ and $f_{2}$ are kept in contact coaxially. The resultant power of combination will be

\frac{f_{1} f_{2}}{f_{1} - f_{2}}

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

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f_{1} + f_{2}

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

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Solution

The correct option is B $\frac{f_{1} + f_{2}}{f_{1} f_{2}}$
When two lenses are placed coaxally then their equivalent focal length $F$ is given as

$\frac{1}{F_{1}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \Rightarrow F = \frac{f_{1} f_{2}}{f_{1} + f_{2}}$

Now power $= \frac{1}{focal length} = \frac{f_{1} + f_{2}}{f_{1} f_{2}}$

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Similar questions

f_{1}

f_{2}

If two thin lenses of focal length f₁ and f₂ are held in contact then the focal length of the combination will be

f_{1}

f_{2}

f_{1}

f_{2}

\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}

∣ ∣ ∣_{1} \times_{2} ∣ ∣ ∣ =_{1} ._{2}

∣ ∣ ∣_{1} +_{2} ∣ ∣ ∣

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