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sha0403

  • 2 years ago

suppose you have a convex lens with focal length 17 cm.if an object is place p cm away from the lens,its image will appear to be q cm from the lens, where p,q and 17 cm are related by the following equation:- 1/17=1/p + 1/q This is known as lens equation. what the rate of change of p with respect to q?

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  1. Shadowys
    • 2 years ago
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    You might want to differentiate this. Did you attempt it?

  2. sha0403
    • 2 years ago
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    differentiate all or what? i don't know..please help me

  3. Shadowys
    • 2 years ago
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    Differentiate the \(\frac{1}{17}=\frac{1}{p}+\frac{1}{q}\) wrt to q. Like this: \(\frac{d(\frac{1}{17})}{dq}=\frac{d(p^{-1})}{dq}+\frac{d(q^{-1})}{dq}\) \(0=-p^{-2}p'-q^{-2}\) So \(p'=- \frac{p^2}{q^2}\) Now sub \(p=(\frac{1}{17}-\frac{1}{q})^{-1}\) into that and you've got it.

  4. sha0403
    • 2 years ago
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    what meaning of p' ?

  5. Shadowys
    • 2 years ago
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    \(p'=\frac{dp}{dq}\)

  6. sha0403
    • 2 years ago
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    yes i got it..thanks a lot..:)

  7. Shadowys
    • 2 years ago
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    You're welcome :)

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