Two convex lenses of focal length $f_{1}$ and $f_{2}$ are mounted coaxially seperated by a distance. If the power of the combination is zero, the distance between the lenses is

| f_{1} - f_{2} |

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| f_{1} + f_{2} |

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\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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Solution

The correct option is B $| f_{1} + f_{2} |$
Let $d$ be the distance between the lens, then

Power $P = \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}} = 0$

$\frac{d}{f_{1} f_{2}} = ∣ ∣ ∣ \frac{1}{f_{1}} + \frac{1}{f_{2}} ∣ ∣ ∣ = ∣ ∣ ∣ \frac{f_{1} + f_{2}}{f_{1} f_{2}} ∣ ∣ ∣$

$d = | f_{1} + f_{2} |$

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Similar questions

f_{1}

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

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

∣ ∣ ∣_{1} +_{2} ∣ ∣ ∣

f_{1}

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

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