Answer
a) $900\times$
b) $1.8\;\rm cm, \;0.3003\;cm$
c) $0.350\dot3\;\rm cm$
Work Step by Step
a) We know that the total magnification is given by
$$M=M_o(M_e+1)$$
since we know that the magnification of the eyepiece is increased by one.
Plugging the known;
$$M=60(14+1)=\color{red}{\bf 900 }\times$$
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b) Thus, the focal length of the eyepiece is given by
$$M_e=\dfrac{N}{f_e}$$
So,
$$f_e=\dfrac{N}{M_e}=\dfrac{25}{14}=\color{red}{\bf1.8}\;\rm cm$$
To find the focal length of the other lens, we need to find the object's position $d_{o2}$;
$$\dfrac{1}{f_e}=\dfrac{1}{d_{o2}}+\dfrac{1}{d_{i2}} $$
$$d_{o2}=\left[ \dfrac{1}{f_e}- \dfrac{1}{d_{i2}}\right]^{-1}$$
Plugging the known;
$$d_{o2}=\left[ \dfrac{1}{1.8}- \dfrac{1}{-25}\right]^{-1}=\bf 1.68\;\rm cm$$
Therefore, the first image is given by
$$d_{i1}=l-d_{o2}$$
where $l$ is the distance between the two lenses.
$$d_{i1}=20-1.68=\bf 18.32\;\rm cm$$
We also know that the image magnification is given by
$$m=\dfrac{d_{i1}}{d_{o1}}$$
Thus,
$$d_{o1}=\dfrac{d_{i1}}{m}=\dfrac{18.32}{60}=\color{red}{\bf\bf0.305\dot 3}\;\rm cm\tag{Part c}$$
Therefore, the focal length of the objective lens is given by
$$\dfrac{1}{f_o}=\dfrac{1}{d_{o1}}+\dfrac{1}{d_{i1}} $$
$$ f_o =\left[\dfrac{1}{d_{o1}}+\dfrac{1}{d_{i1}} \right]^{-1}$$
Plugging the known;
$$ f_o =\left[\dfrac{1}{0.305\dot 3}+\dfrac{1}{18.32 } \right]^{-1}=\color{red}{\bf0.3003}\;\rm cm$$
