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anonymous
 4 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?
anonymous
 4 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You might want to differentiate this. Did you attempt it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0differentiate all or what? i don't know..please help me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Differentiate 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i got it..thanks a lot..:)
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