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sha0403
Group Title
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?
 one year ago
 one year ago
sha0403 Group Title
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?
 one year ago
 one year ago

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Shadowys Group TitleBest ResponseYou've already chosen the best response.1
You might want to differentiate this. Did you attempt it?
 one year ago

sha0403 Group TitleBest ResponseYou've already chosen the best response.0
differentiate all or what? i don't know..please help me
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

sha0403 Group TitleBest ResponseYou've already chosen the best response.0
what meaning of p' ?
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
\(p'=\frac{dp}{dq}\)
 one year ago

sha0403 Group TitleBest ResponseYou've already chosen the best response.0
yes i got it..thanks a lot..:)
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
You're welcome :)
 one year ago
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