Question
Show that for thin lenses that have focal lengths $f_{1}$ and $f_{2}$ and are in contact, the effective focal length $(f)$ is given by$$\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$$
Step 1
The focal lengths of the lenses are given as $f_1$ and $f_2$. Show more…
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Two thin lenses of focal lengths $f_{1}$ and $f_{2}$ are in contact. Show that they are equivalent to a single thin lens with $$f=\frac{f_{1} f_{2}}{f_{1}+f_{2}}$$ as its focal length.
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Show that if two lenses of focal lengths $f_{1}$ and $f_{2}$ can be considered to have zero physical separation, then the effective focal length of the combination of lenses is $$\frac{1}{f_{\mathrm{eff}}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$$ Note: Assuming that the actual physical separation of the lenses is $x$, this approximation is strictly valid only when $f_{1} \gg x$ and $f_{2} \gg x$
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