I have just started learning optics at school and my teacher derived the lens and mirror formulas. While doing so, she applied the sign convention for $u$, $v$ and $f$ and arrived at the final expression. However, while solving problems using the mirror formula why do we again apply sign convention for the given values of $u$, $v$ or $f$?


3 Answers 3


Here's what you did in the derivation actually.

You derived the lens law for a special case. i.e. for a particular position of the lens, object and image. Like image on right of the lens and object on left or vice-versa.

Now the lens formula that you got was for this particular scenario. Now suppose you apply the sign convention on this formula once again, the sign convention you applied in the derivation will be cancelled.

It is like if 'u' is negative in your formula, applying another '-' will cancel the effect and give you the general lens formula that will hold in all situations.

Its kind of like applying sign convention gives a particular formula and applying it twice cancels the effect of the sign convention at all,giving you the general lens formula.

This general lens formula is what you were after which you can use in other special scenarios by applying the sign convention.

If you want to see what I am saying, try deriving the lens formula, using these two conditions.

(1) image on right and object on left of lens. (2) image on left and object on left.

BTW, do you know why you use sign convention in the first place ? Because the formula can't distinguish otherwise between right and left, as you primarily using only magnitude of lengths in the derivation through similar triangles.

Also, do you know a virtual image is kind of an optical illusion only, as to there is nothing there behind the mirror.


Maybe this is a bit too general an answer but I hope this helps someone in the future. Here is the lenses makers equation write explicitly with the sign conversion terms,

$$ \frac{1}{f} = \left(n-1\right)\left( \frac{d\left(n-1\right)}{\beta_1\beta_2n\bigl|R_1\bigr|\bigl|R_2\bigr|} + \frac{\beta_1}{\bigl|R_1\bigr|} - \frac{\beta_2}{\bigl|R_2\bigr|} \right) $$

This way you can ignore the sign of the lens radius and simply set the sign terms ($\beta_1$ and $\beta_2$) to correct values for the particular type of lens you are studying.

The sign convention I have applied is:

  • $\beta_1=+1$ when left surface is convex
  • $\beta_1=-1$ when left surface is concave
  • $\beta_2=+1$ when right surface is concave
  • $\beta_2=-1$ when left surface is convex

If you want to calculate radii to give fixed focal length then having the equation in this form makes rearranging for the required terms very simple.

  • $\begingroup$ please define your variables $\endgroup$ Commented Nov 24, 2020 at 22:59

There are different sign convention in optics. Once you accept one sign convention you should make sure that the answer is in the same convention.

Application of the signs twice guarantee the same. How this magic happening?

At the time of derivation you applaying the signs for U,V,F thar guarantee that U,V,F are in our convention. At the time of solving problems you apply signs for only two variables. the effect of signs of that variable cancells. But for the third, RESULT, you only applaying convention once that at the time of derivations and that remains there which ensure that the answer is in the same convention.


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