# Reflection at a spherical surface

1. May 16, 2014

### Jon.G

1. The problem statement, all variables and given/known data
An object 0.6cm tall is placed 16.5cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22.0cm.
Determine the position, size, orientation and nature of the image.

2. Relevant equations
$\frac {1}{S} + \frac {1}{S'} = \frac {1}{f} = \frac {2}{R}$

3. The attempt at a solution
$\frac {1}{16.5}+ \frac {1}{S'} = \frac {2}{-22}$
$\frac {1}{S'} = -\frac{1}{11} - \frac {1}{16.5}$
$S' = -6.6cm$
However, the answer given is 33.0cm, which you get when R = +22.0 , not -22.0
In my lecture notes is definitely says that the radius of curvature for a concave mirror is negative.
Are the notes wrong? Or is there something I am missing?

(Using 33.0cm as S' I can get the answers to the other parts of the question, I just want a little bit of help in understanding why the radius of curvature is positive)

Thanks

2. May 16, 2014

### ehild

The focal length is taken positive in case of a concave mirror. F=|R|/2

ehild

3. May 16, 2014

### dauto

The notes are wrong. Concave mirrors and convex lenses have positive focal distance while convex mirrors and concave lenses have negative focal distances.

4. May 16, 2014

### haruspex

I'm not sure that there is a universally agreed convention. The important thing is to pick a convention and ensure all your equations conform to it.

The convention used at http://en.wikipedia.org/wiki/Radius_of_curvature_(optics) and http://en.wikipedia.org/wiki/Focal_length seems good to me. You treat the incoming ray as coming from the negative side to the positive. The radius of the surface is always taken as the offset from the surface to the centre of curvature. So we have concave negative, convex positive whether it be a mirror, or the entry surface of a lens, but reversed for the exit surface of a lens.
Correspondingly, the focal length formula for a lens uses 1/R1 - 1/R2. Thus, for a biconvex lens we get a positive minus a negative, and the curvatures reinforce.

The error in the OP is the sign of S.

5. May 17, 2014

### dauto

Yes there are alternative sign conventions but the one I described is the easiest.

6. May 17, 2014

### haruspex

How so?

7. May 18, 2014

### dauto

Gaussian sign convention is easier for students that are being introduced to the subject because
* The equations used for mirrors and lenses become identical
* A minus sign for image location corresponds to a virtual image and the same thing is true for an object

* The sign convention for an object seems backwards (the left sign of the axis is negative, they are not used to that)

My experience is that students find the advantages to outweigh the disadvantages