Two lenses of focal lengths $f_{1}$ and $f_{2}$ are placed side by side. What is the focal length of the composite lens?

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

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f_{1} + f_{2}

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Solution

The correct option is A $\frac{f_{1} f_{2}}{f_{1} + f_{2}}$

The net power of lenses of individual powers $P_{1}, P_{2}, P_{3},$ and so on, placed in contact is given by,

Net power, $P = P_{1} + P_{2} + P_{3}$ ...

The power $P_{1}$ of the lens of focal length $f_{1}$ = $\frac{1}{f_{1}}$ .

The power $P_{2}$ of the lens of focal length $f_{2}$ = $\frac{1}{f_{2}}$ .

$P = P_{1} + P_{2}$ = $\frac{1}{f_{1}} + \frac{1}{f_{2}}$
Let $f$ be the focal length of the combination.

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

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

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Similar questions

f_{1}

f_{2}

Two thin lenses of focal lengths $f_{1}$ and $f_{2}$ are placed in contact with each other. The focal length of the combination will be given by:

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} a n d f_{2}

d < f_{1}, d < f_{2}

Δ

^{'} f^{'}

^{'} f^{'}

\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}}

F =

f_{1}

f_{2}

^{'} d^{'}

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