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I have a problem understanding the sign conventions used in the equations that describe the imaging properties of mirrors - in particular, when and why should I use a positive or negative number for the object and image distance when both are on the same side?

I would like to illustrate my confusion with an example:

A dime $36.5$ cm away from, and on the optical axis of, a concave spherical mirror produces an image on the same side of the mirror as the source. The image is $11.7$ cm away from the mirror. If the dime is moved on the axis to $21.7$ cm from the mirror, how far away from the mirror is the image now?

Here I used the object distance $o$ to be negative.

$$o_1 = -.365$$

$$o_2 =\quad ?$$

$$i_1 = .117$$

$$i_2 = .217$$

My attempt:

$$\frac{1}{i} + \frac{1}{o} = \frac{1}{f}$$

Solve for $\dfrac{1}{f}$: $(\frac{1}{i_1} + \frac{1}{o})$

$$\frac{1}{(\frac{1}{f} - \frac{1}{i_2})} = o_2 = 15 \:\mathrm{cm}$$

But wait. Below I didn't use the object distance $o$ to be negative. In fact for this and several problems I used it to be positive. It comes out correct, and incorrect if i use the object distance $o$ to be negative. In fact I wanted to use the image distance to be negative, because it's going to the left of the mirror instead of the right. (It's concave.)

A candle is placed $16.5$ cm in front of a thin converging lens of focal length $4.7$ cm. What is the image distance? ( Note, an inverted image will have a "negative" size.)

$$\frac{1}{o} + \frac{1}{i} = \frac{1}{f}$$

$$o = .165$$

$$f = .047$$

solve for $i$: $\frac{1}{ \frac{1}{f} - \frac{1}{o}} = 6.57$ cm

Secondly, how do I find the radius of curvature of the first part?

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  • $\begingroup$ Any lens or mirror equation should come with a set of rules specifying exactly what sign conventions apply for real and virtual images, concave or convex surfaces for that version of the formula... $\endgroup$ – DJohnM Nov 18 '14 at 3:17
  • $\begingroup$ Please take some time to read our guidelines on homework and exercise questions. You do have a real conceptual question there, but you should rephrase your post in a way which makes it maximally useful for other people that are not staring at the precise same problem as you. $\endgroup$ – Emilio Pisanty Nov 18 '14 at 12:35
  • $\begingroup$ @EmilioPisanty - I have edited the question along the lines you suggested. Think it can stay open now? $\endgroup$ – Floris Nov 18 '14 at 14:50
  • $\begingroup$ The introduction is better but the end just trails off into numbers. What do you find confusing about them? Elaborate on the concepts, and only use the numbers as a reference. $\endgroup$ – Emilio Pisanty Nov 18 '14 at 14:59
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/70730/2451 and links therein. $\endgroup$ – Qmechanic Apr 8 '15 at 16:07
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For a mirror, I’ve found the best convention for signs in optics to be the following.

A distance to image $s$ is negative if it is on the opposite side of the mirror from that which the light originated (i.e., a virtual image). (relative to the mirror - this is important for multi-mirror systems)

A distance to image $s$ is positive if it is on the same side of the mirror as that which the light originated (i.e., a real image). (again, relative to the mirror).

For lenses, the distance to image $s$ is negative if it is on the same side of the lens from that which the light originated… etc. and the opposite for positive.

With these conventions, you can rather easily solve for the focal length:

$\frac{1}{o_1} + \frac{1}{o_2} = \frac{1}{f}$,

$\left[\frac{1}{0.365}+\frac{1}{.117}\right]^{-1} = f$,

and we know that the radius of curvature is given by: $f=\frac{R}{2}$. Thus,

$R = 2\left[\frac{1}{0.365}+\frac{1}{.117}\right]^{-1}$.

Now, if the object is moved to a new distance 21.7cm, we have:

$\frac{1}{.217} + \frac{1}{o_2} = \frac{1}{f}$,

and we now know $f$ from our above calculation. The rest is simple.

[EDIT: For the record, notice that my convention is consistent with your observation that you got the correct answer in the second case with a negative image coordinate… you are in fact correct, in that regard. :) ]

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