Answer
(a) $4.10mm $
(b) $-8.0cm $
Work Step by Step
(a) We know that
$\frac{1}{d_{\circ}}=\frac{1}{f_{obj}}(1+\frac{1}{m})$
We plug in the known
$\frac{1}{d_{\circ}}=\frac{1}{4.00mm}(1+\frac{1}{-40.0})$
$\frac{1}{d_{\circ}}=\frac{0.975}{4.00mm}$
$\frac{1}{d_{\circ}}=0.243$
$\implies d_{\circ}=4.10mm $
(b) We can find the required focal length as follows:
$ f_e=\frac{m_{\circ}N}{M_{total}}$
We plug in the known values to obtain:
$ f_e=\frac{(-40.0)(25cm)}{125}$
$ f_e=-8.0cm $
