Here is an example of Cassegrain telescope: Parallel rays from a distant object get reflected by the concave mirror forming an image at its focus behind the convex mirror. This image acts as a virtual object for the convex mirror, and it forms a real image in front. Let's apply mirror equation for this convex mirror to find the distance at which final image is formed.enter image description here

1/v + 1/u = 1/f

u = + 90; f= +70;

Therefore, 'v' comes out to be +315 cm. I am troubled to see this positive sign in image's distance, would not this mean that the image is formed to the left of convex mirror? Does sign convention fail in this case? Or have I made some fatal mistake?

The exact question statement goes like this:

enter image description here

  • 1
    $\begingroup$ The latter :-) . Look up similar mirror configurations in any optics text (Smith, Modern Optical Engineering comes to mind) and you'll see where you got propagation sign errors. $\endgroup$ – Carl Witthoft Jun 20 '14 at 15:54
  • $\begingroup$ Thanks Carl. Let me find this book. I still don't understand what's the fatal error, how I should incorporate primary's result in secondary. $\endgroup$ – Swami Jun 21 '14 at 2:37
  • $\begingroup$ Help me Carl! I found an e-copy of this book, but could not understand anything, there is no direct reference to the usage of sign convention for multiple reflection problems. Would you be so generous as to spare some time for this particular problem? Much respect! Thanks! $\endgroup$ – Swami Jun 21 '14 at 3:22

For the calculation you are doing, the primary mirror is irrelevant. You used the primary properties to determine where in space the first, real (for the primary) image is created.

You then shift gears, and realize that this image is a virtual image , with respect to the secondary at its defined position So when you apply the mirror equation for the second time, the solution you get implies a real image for the secondary mirror, located in front of the secondary mirror...

Incidentally, there are many versions of the mirror formula, with different signs between the terms. Each comes with its own convention as to which positions and radii are positive or negative. Much confusion ensues...


For example, if we apply this formula with its conventions (from Wikipedia): enter image description here

to the secondary mirror problem, the object distance and focal length would be negative...

| cite | improve this answer | |
  • $\begingroup$ How should I proceed? Thanks anyways. $\endgroup$ – Swami Jun 21 '14 at 2:35
  • $\begingroup$ That's right! And that is my problem. Object's distance, and focal length is negative, which would make image's distance negative, which would in turn mean image is formed to the left of secondary. Ain't it? $\endgroup$ – Swami Jun 21 '14 at 3:25
  • 1
    $\begingroup$ Where did the diagram come from? What dimension does the "90 cm" refer to? It's only "110 cm" from the primary to the prime focus! $\endgroup$ – DJohnM Jun 21 '14 at 14:29
  • 1
    $\begingroup$ Then the diagram is seriously distorted...Again, is there a source for the data in this problem? $\endgroup$ – DJohnM Jun 22 '14 at 2:27
  • 1
    $\begingroup$ Interesting... Google Books (searched for "uses two mirrors as shown in") lists four different texts that use the question exactly as posted, but without the Figure number. Several show the solution; the solutions are the same, and the solution is wrong. Your solution, posted in the original question, is correct, and the system as described does not form a real image. Congratulations... $\endgroup$ – DJohnM Jun 22 '14 at 4:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.