Question

Please assist.

Answer 1

can you answer my question next?

Answer 2

Suppose that an object has thickness \(du\) so that it extends from object distance \(u\) to \(u + du\). Prove that the thickness \(dv\) of the image is given by \(- \frac{v^2}{u^2}du\), so the longitudinal magnification is given by \(\frac{dq}{dp} = -M^2\) where \(M\) is the lateral magnification.

Answer 3

me next

Answer 4

\[\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}\]\[\Rightarrow -\dfrac{1}{u^2} - \dfrac{1}{v^2}\dfrac{dv}{du} = 0\]\[\Rightarrow \dfrac{dv}{du} = - \dfrac{v^2}{u^2}\]

Answer 5

Oooyeah.