# Sign convention in optics

Why is the sign convention used in the derivation of the lens formula and yet used again when it is applied in numerical problems? Won't the whole idea of sign convention be eliminated if it is used twice?

If you're calculating something like "Where is the image in this lens setup?", there shouldn't be any "convention" in the final answer.

"The image is right here, where I'm pointing" $\leftarrow$ that statement is objectively true or false, it cannot depend on a convention.

Therefore any conventions you use in the process of calculating this thing should not affect the final answer. That's why the sign conventions tend to appear twice: Once in the definition of a quantity, and once in the formula related to that quantity. The two uses cancel out in the final answer. Which is as it should be.

There are actually two different sign conventions in optics. Without any convention it's hard to develop any universal scientific statement (like formula) so let's make up one here:

F - lens focal length, d - lens to object distance, s - lens to image distance.

1. Cartesian sign convention

Small lens formula:

1/s - 1/d = 1/F

Interpretation: "curvature after - curvature before = curvature added". This convention is widely accepted by professional opticians as it employs idea of "positive = co-propagating with left-to-right ray direction".

2. "Classic" scholar sign convention

Small lens formula:

1/s + 1/d = 1/F

s>0 and d>0 if image and objects are real and both negative if both are virtual.

3. Without any sign convention you would have to use formula like this one:

± 1/s ± 1/d = ± 1/F

Depending on the type of lens and object position:

• Convex lens +1/F, Concave lens -1/F;
• Real image +1/s, virtual image -1/s;
• Diverging ray fan (real object) +1/d, converging ray fan (virtual object) -1/d.
• What is the relationship between convex/concave and sign of f in the first two equations? – DJG Oct 7 '19 at 17:04

The sign convention is important in how you define your quantities. So the thin lens formula equation (found here):

$\frac 1 f= \frac 1 {s'}+\frac 1 s$

This will only give you correct results, if you follow the correct sign conventions. In this case you take the lens to be 0, and then, anything to the left is negative (normally the object), and anything to the right is negative. This is also important in terms, if you have a positive or negative lens. You have to use the same sign convention to keep your derivation consistant.

• Anything to the left is negative, and anything to the right is also negative? Hmm... – DJG Oct 7 '19 at 17:05

Actually in lens maker's formula derivation we apply sign convention(Cartesian sign convention) twice.Thus it gets cancelled in derivation.So in the problem, when we again apply sign convention we get the right answer in accordance with sign convention.

But personally for me Classical sign convention is much easier to follow and solve problems. I use following sign convention for lens: i) Distances of real object, real image and real focus (focal length) are positive, if not they are negative. ii) If principal focus is in the denser medium then focal length is positive. iii) If the incident ray bends towards the principal axis after refraction, then the image is real and if it bends away then the image is virtual.

Yes, it will, that is the whole idea. The reason is that everybody has to get the same physical answer independent of the sign convention.

The problem is that you are free to choose positive radius of curvature for convex or concave surfaces, positive focal lengths for a converging or diverging lens, and so on, so, conventions are necessary standards within a text or community.