Magnification of closely packed thin lenses, or of closely packed lens and mirror

Question

I was taught how to calculate the equivalent foci in both the cases. And since the formulae resemble the simple mirror and lens maker formulae, teacher said that this system is behaving like a single mirror/lens. It made sense as long as we calculated the image or object distance using those. But when it came to magnification after that, he simply did v/u or (-v/u), putting in the values of final image coordinate and initial object coordinate, and gave the reason that, this is doable because the system is acting like a single lens/mirror.

What came to my mind though, is that we are basically taking the image of first lens as the object of the second lens (or the first mirror), right? So the magnification overall should be the product of the individual magnifications, and cannot be calculated as he did, and I found myself correct according to a textbook I use. Now I'm confused.

Answer 1

In general you are correct, but in the special case where the lenses are very close together your teacher is correct as well. Consider two lenses placed so close together that we can approximate them as being in the same place:

The object is a distance $u_1$ from the first lens and if we take the first lens on its own then it forms an image at $v_1$, and the magnification for this step is:

$$ M_1 = \frac{v_1}{u_1} $$

Then we take the image from the first lens as a virtual object for the second lens, and the second lens forms an image at $v_2$. The magnification is:

$$ M_2 = \frac{v_2}{u_2} $$

As you say, the total magnification is the product $M_1M_2$:

$$ M = M_1M_2 = \frac{v_1}{u_1}\frac{v_2}{u_2} $$

But because the lenses are in the same place $u_2 = v_1$ so they cancel in the fraction and we are left with:

$$ M = M_1M_2 = \frac{v_2}{u_1} $$

just as your teacher said!

But note that we only have $u_2 = v_1$ because both lenses are (approximately) at the same place. If the spacing between the lenses was larger this would no longer be true and your teacher's approximation would no longer work while your method of multiplying the magnifications would still be fine.