Q.A compound microscope has an objective of focal length $1.5\ \text{cm}$ and an eyepiece of focal length $5\ \text{cm}$, with the two lenses separated so that the objective forms its image at a distance $L=20\ \text{cm}$ from the eyepiece. If the final image is formed at the near point ($D=25\ \text{cm}$) of a normal eye, calculate the magnifying power of the microscope using $M\approx\dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)$.

The objective forms a real, inverted, magnified image just inside the eyepiece's focal length, which the eyepiece further magnifies as a simple magnifier; total $M=M_0\times M_e$.

1. Ray diagram (described): The object $O$ is placed just beyond the focus $F_0$ of the objective lens. The objective forms a real, inverted, highly magnified image $A_1B_1$ a little inside the focal length $f_e$ of the eyepiece. This intermediate image acts as the object for the eyepiece, which — since $A_1B_1$ lies within $f_e$ — acts as a simple magnifier and produces a virtual, further magnified, final image $A_2B_2$, inverted with respect to the original object, seen at or beyond the near point.
2. Magnifying power: Overall magnification is the product of the objective's linear magnification and the eyepiece's angular magnification: …