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## sha0403 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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1. Shadowys

You might want to differentiate this. Did you attempt it?

2. sha0403

differentiate all or what? i don't know..please help me

3. Shadowys

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

what meaning of p' ?

5. Shadowys

$$p'=\frac{dp}{dq}$$

6. sha0403

yes i got it..thanks a lot..:)

7. Shadowys

You're welcome :)