# Suppose an object has thickness dp so that it extends from object distance p to p + dp.

Suppose an object has thickness dp so that it extends from object distance p to p + dp.

(a) Prove that the thickness dq of its image is given by (−q2/p2)dp. (Do this on paper. Your instructor may ask you to turn in this work.)

(b) The longitudinal magnification of the object is Mlong = dq/dp. How is the longitudinal magnification related to the lateral magnification M? (Use the following as necessary: M.)
Mlong =
asked Apr 11, 2012 in Physics

Part A:

1/p + 1/q = 1/f

since 1/f is a constant we can rewrite as:

1/p + 1/q = constant

or rewritten as:

p^-1 + q^-1 = constant

Now take the derivative with respsect to p.

-1p^-2 + -q^-2 *dq/dp = 0

Using algebra this can be rearranged to the desired result of:

dq = - (q^2 / p^2) * dp

Part B:

M = - q/p

Mlong = dq/dp

From part A, we can solve for dq/dp

which results in:

dq/dp = - (q^2 / p^2)

(q^2 / p^2) can be replaced by M^2,

so we get

Mlong = - M^2

answered Apr 14, 2012 by ~Expert~ (3,020 points)
selected Apr 14, 2012 by awesome
why arenot apply sign convention in this proof ?