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Two thin biconvex lenses have focal lengths $f_{1}$ and $f_{2}$. A third thin biconcave lens has focal length of $f_{3}$. If the two biconvex lenses are in contact, then the total power of the lenses is $P_{1}$. If the first convex lens is in contact with the third lens, then the total power is $P_{2}$. If the second lens is in contact with the third lens, the total power is $P_{3}$, then
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$P_{1}=\frac{f_{1}+f_{2}}{f_{1} f_{2}}, P_{2}=\frac{f_{3}-f_{1}}{f_{1} f_{3}}$ and $P_{3}=\frac{f_{3}-f_{2}}{f_{2} f_{3}}$
According to cartesian sign conversion,
Focal length of first biconvex lens $=f_{1}$
Focal length of second biconvex lens $=f_{2}$
Focal length of third biconcave lens $=-f_{3}$
$\therefore$ Focal length of combination of first and second lenses is
$\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{f_{1}+f_{2}}{f_{1} f_{2}}$
As, power of lens, $P_{1}=\frac{1}{f}=\frac{f_{1}+f_{2}}{f_{1} f_{2}}$
Similarly, power of combination of first and third lenses is
$P_{2}=\frac{1}{f}=\frac{1}{f_{1}}-\frac{1}{f_{3}}=\frac{f_{3}-f_{1}}{f_{1} f_{3}}$
and for combination of second and third lenses is
$P_{3}=\frac{1}{f}=\frac{1}{f_{2}}-\frac{1}{f_{3}}=\frac{f_{3}-f_{2}}{f_{2} f_{3}}$
Focal length of first biconvex lens $=f_{1}$
Focal length of second biconvex lens $=f_{2}$
Focal length of third biconcave lens $=-f_{3}$
$\therefore$ Focal length of combination of first and second lenses is
$\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{f_{1}+f_{2}}{f_{1} f_{2}}$
As, power of lens, $P_{1}=\frac{1}{f}=\frac{f_{1}+f_{2}}{f_{1} f_{2}}$
Similarly, power of combination of first and third lenses is
$P_{2}=\frac{1}{f}=\frac{1}{f_{1}}-\frac{1}{f_{3}}=\frac{f_{3}-f_{1}}{f_{1} f_{3}}$
and for combination of second and third lenses is
$P_{3}=\frac{1}{f}=\frac{1}{f_{2}}-\frac{1}{f_{3}}=\frac{f_{3}-f_{2}}{f_{2} f_{3}}$
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