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Whilst finding the magnification of a compound microscope, we multiply the linear magnification of the objective with the angular magnification of the eyepiece. This gives us the angular magnification of the microscope.

Why do we use angular magnification for the eyepiece and not linear magnification? Or alternatively, why don't we use the angular magnification of the objective?

I read the following similar question answered here a couple of years ago - Magnification in compound microscope

But I haven't learnt about wave optics so how do you explain this using ray optics?

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Angular magnification is used when linear magnification is not defined. This is the case if an object or an intermediate image is in the focal plane of considered lens (for the compound microscope: the eyepiece). According to the thin lens formula $$\frac{1}{f}=\frac{1}{S_o}+\frac{1}{S_i}$$ if $S_o\to f$ we have $S_i\to \infty$. But if the image is created at infinity with infinite size (which is equivalent to the image rays of one object point all being parallel), there is no sense in talking about linear magnification.

The eyepiece is usually considered as being adjusted so as to view with the relaxed eye, which exactly corresponds to the requirement $S_i\to \infty$ (at least for normal-sighted people). The final image has of course finite size, because it is created as a real image on your retina. But since this image depends on the specific properties of the viewer's eye, this can't be used explicitly.

On the other hand, the objective creates a real intermediate image, which can be expressed as a linear magnification.

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    $\begingroup$ Thank you so much! I was struggling with this for hours. Got it now :) $\endgroup$
    – Ash
    Apr 5 at 15:22
  • $\begingroup$ My pleasure! :-) $\endgroup$
    – oliver
    Apr 5 at 15:22

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