Question

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?

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

what meaning of p' ?

Answer 5

\(p'=\frac{dp}{dq}\)

Answer 6

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

Answer 7

You're welcome :)