Can we change the sign convention (distance measured in the direction of incident ray positive and opposite negative) midway solving a problem. According to me we should ,as the mirror/lens formula is for an individual apparatus,and cannot be for the whole problem.


An example of the above concept enter image description here

enter image description here In this arrangement, a ray of light Is first refracted by lens L. then at reflected at curved mirror M. and finally refracted once again at lens L. Let object O be located in front of the lens. Let the image from lens $l_1$, be formed at $v_1$Then. from the lens- formula we have $\frac{1}{v_1} -\frac 1u =\frac{1}{f_l1}$ (i)

where $f_l1$ is. the focal length Of the Iens L when the ray approaches from the left. $I_1$will now be the object for mirror M. Let image $I_2$ be formed after reflection at $v_2$ from the pole of the mirror. Then.

$\frac{1}{v_2} + \frac{1}{v_1} = \frac{1}{f_m}$. (ii)

where $f_m,$ is the focal length of mirror M. Now, $I_2$ will be the object for the final refraction at lens L. If $I_3$ be the final image formed at v from the center of the lens,

Then we have . (iii) $\frac 1v - \frac{1}{v_2} = \frac{1}{f_l2}$

where fl2. is the focal length of the lens, L when the ray of light approaches from the right. When a ray of light approaches from the left. the positive X-axis is from Is from left to right.

Is it were replaced by a concave mirror today focal length $f_e$ then

$\frac 1v + \frac 1u =\frac{1}{f_e}$

On solving we get $f_e =f_m$

This is definitely wrong.

I think problem is somewhere with the sign convention in the last refraction. As for last refraction coordinate system cannot be same as in the first refraction and reflection convention is positive towards right. So what will be the new mirror formulae. By hit and trial it comes out to that simply replace f>-f But according to me if convention is reversed ( right is negative and left is positive ) lens formula becomes. $-\frac 1v - (-\frac 1u) =- \frac 1f $

Which is actually the same formulae ?? I am terribly confused in such a simple problem .

Thanks in advance


closed as off-topic by John Rennie, user259412, Jon Custer, Daniel Griscom, JamalS Oct 9 '17 at 6:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, user259412, Jon Custer, Daniel Griscom, JamalS
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  • $\begingroup$ Hi Varang and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Oct 1 '17 at 6:57
  • $\begingroup$ You should find equivalent power instead of focal length as power of converging system,either convex lens or concave mirror,has same sign. $\endgroup$ – Rajendra Pd Oct 1 '17 at 7:17
  • 1
    $\begingroup$ I think this question doesn't run afoul of our homework policy, since it seems to be asking a conceptual question about why a derivation doesn't work, not asking for the answer to the homework question. That being said, it would probably benefit from some editing to clarify that. $\endgroup$ – David Z Oct 1 '17 at 7:30
  • $\begingroup$ @JohnRennie :It isn't a worked type problem. I came up with the above derivation on my own and it will definitely help somebody looking for new lens formula when sign convention is not followed. ;) $\endgroup$ – varang rai Oct 1 '17 at 7:40