# Problems dealing w/ mirrors

1. May 31, 2005

### chase222

I have 2 questions regarding mirrors:

If you look at yourself in a shiny Christmas tree ball with a diameter of 9.0 cm when your face is 30.0 cm away from it, where is your image? Is it real or virtual? Is it upright or inverted?

I know the mirror equation is:

(1/object distance) + (1/image distance) = 1/f
I know that the object distance is 30.0 cm, but I am confused as to what the 9.0 cm represents. Do I divide that number in half to equal 4.5, and would that number be my radius of curvature? Also I don't understand how you tell if an image is real or virtual, and upright or inverted? Is it just inverted if the number is negative?

The other problem is:

How far from a concave mirror (radius 27 cm) must an object be placed if its image is to be at infinity?

To solve this I set up the mirror equation from above, except I rearranged it to be:
(1/f) - (1/image distance) = (1/object distance)
so to plug it in it would be:
(1/13.5)-(1/infinity)=1/object distance
I got the 13.5 by dividing the radius in half. However, I don't get how to solve this problem b/c you can't plug infinity into the calculator.

2. Jun 1, 2005

### Chi Meson

first question. the focal length of any mirror is 1/r. You have been given the diameter, so now you need 1/2 the radius.

Also, all convex mirrors are diverging mirrors, so therfre the focal lengths are defined to be negative quantities. All images in a diverging mirror will always be virtual, upright, and reduced in size.

Second question: 1/infinity = zero