# Radius of Curvature, Lenses, Finding Di

1. Oct 13, 2012

### PeachBanana

1. The problem statement, all variables and given/known data

If you look at yourself in a shiny Christmas tree ball with a diameter of 8.1 cm when your face is 35. 0cm away from it, where is your image?

2. Relevant equations

1/do + 1/di = 1/f

3. The attempt at a solution

1/di = 1/4.05 cm - 1/35.0 cm

di = 4.6 cm

I thought f = 8.1 cm / 2

2. Oct 13, 2012

### Ibix

Not quite - where did you get that formula from?

Also, you need to watch your signs here. You are applying a lens formula to a mirror, and that means some careful thought is needed. Far and away the easiest thing to do is draw a couple of ray diagrams.

First, put the object on the left and the mirror in the middle. Rays go left-to-right, strike the mirror, and go back right-to-left. Add back-tracks to the reflected rays if that's necessary to find the image position.

Second, flip the reflected rays and their back-tracks (if any) around. So, if a ray strikes the mirror and bounces back at 10° above the horizontal, make it 10° above the horizontal in the opposite direction. You have replaced the mirror with a lens that (except for transmitting the light rather than reflecting it) does the exact same thing as the mirror. Now you can apply your lens formula (think about the sign on f!) to this diagram. That should let you label your original diagram easily enough.

3. Oct 13, 2012

### PeachBanana

Hello. The book said the focal point = radius of curvature / 2 but I'm sure I misapplied it. I will go back and do what you said.

4. Oct 13, 2012

### Ibix

Hopefully the book says focal distance, but otherwise I think you've just made a minor slip - 8.1cm is the diameter of the ball, not the radius.

I'd still suggest drawing it out to make sure you get the right sign on f.