The focal length of the objective and eye lenses of a microscope...

Here (v_O=L-f_e=21.7 \mathrm{ cm}-2.5 \mathrm{ cm}=+19.2 \mathrm{ cm}) $+v e$ sign shows that image is real. For a objective lens (\frac{1}{u_O}+\frac{1}{v_O}=\frac{1}{f_O}) [using real-virtual sign convention] $$\begin{aligned} & \therefore \quad \frac{1}{u_O}=\frac{1}{f_O}-\frac{1}{v_O} \quad \therefore \quad \frac{1}{u_O}=\frac{1}{(+1.6)}-\frac{1}{(+19.2)} \\ & \frac{1}{u_O}=\frac{19.2-1.6}{(1.6)(19.2)} \quad \text{\text{ or } } \quad u_O=+1.745 \mathrm{ cm} \end{aligned}$$) +ve sign shows that object is real. When the image is formed at infinity, the magnification is given by (m=\left(\frac{v_O}{u_O}\right)\left(\frac{D}{f_e}\right)) where $D$ is the distance of distinct vision. (\therefore \quad m=\left(\frac{19.2 \mathrm{ cm}}{1.745 \mathrm{ cm}}\right)\left(\frac{25 \mathrm{ cm}}{2.5 \mathrm{ cm}}\right)=110 .)$