# Sign convention in lenses and mirrors [closed]

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.

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An example of the above concept

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 .