Answer
a) $f_2: objective, f_1: eyepiece$
b) $M=7.84$
c) $f_1: objective, f_2: eyepiece$
d) $M=-12$
Work Step by Step
(a) We know that in the given scenario, the lens with more focal length should be used as the objective and the lens with less focal length should be used as the eye piece. Thus, $f_2=20.4cm$ should be used as the objective and the lens with $f_1=2.60cm$ should be used as the eye piece.
(b) The maximum angular magnification is given as
$M=\frac{f_{objective}}{f_{eye \space piece}}$
$\implies M=\frac{20.4cm}{2.60cm}=7.84$
(c) In this case, the lens with more focal should be used as the eye piece and the lens having less focal length should be used as the objective. Thus, $f_2=20.4cm$ is used as the eye piece and $f_1=2.60cm$ is used as the objective.
(d) We know that
$d_i=(\frac{1}{f_{objective}}-\frac{1}{d_{\circ}})^{-1}$
$\implies d_i=(\frac{1}{2.60cm}-\frac{1}{2.90cm})^{-1}=25.13cm$
Now $M=-\frac{d_i N}{f_{objective} f_{eyepiece}}$
$\implies M=-\frac{(25.13cm)(25.0cm)}{(2.60cm)(20.4cm)}=-11.846\approx -12$
