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I don't agree with this. Mirror ray diagrams have the virtue that
you're always dealing with real light rays traveling in the direction
of the propagation of the light.
Thus, even if the image from the first mirror is a virtual
image and resides behind the mirror (where there's really no
light at all), it doesn't matter conceptually because the
light rays you are dealing with have bounced off that mirror
and are heading towards the next mirror in an easily
understood direction -- they're coming from the direction of
that virtual image.
But in the case of lenses, one runs into a conceptual problem when
trying to draw ray diagrams that you don't run into with mirrors. I
think that you and John Denker are sweeping this issue under the rug
in your reply. Maybe it would help to give a specific lens
configuration:
When a lamp is placed 42 cm to the left of a particular convex lens,
suppose a real (inverted) image is formed 37.5 cm to the right of the
lens. The lamp and convex lens are kept in place while a concave
lens is mounted 15 cm to the right of the convex lens. A real image
of the lamp is now formed 35 cm to the right of the concave lens.
What is the focal length of each lens?
Applying the lens equation to the convex lens, one obtains +19.8 cm.
Applying the lens equation again to the concave lens (using the image
of the convex lens as the object of the concave lens), one obtains
-63 cm.
But one runs into a conceptual issue (which I tried to explain
in my first query) if you try to draw the ray diagrams for
this situation. Try it!
The only answer I've come up with is that you can't solve a problem
by drawing ray diagrams when the object of a lens is a virtual object
-- that is to say, the object is on the same side of the lens in
which the light is headed.
To put it another way, you can only draw the diagrams using
rays which are headed in the real direction of the propagation
of light.
That conclusion may seem obvious, but I've never heard that
rule nor seen it in any textbook.