# Homework Help: Curved Mirror mathematics?

1. Jul 17, 2010

### i_m_mimi

What is the formula for mirrors and lenses?
My textbook says 1/f = 1/ do + 1/di
Then, hi/ho = - di/do
Then, M= hi/ho
M = di/do
Why is that that M equals each other, but one di/do is negative and the other isn't??
Is the formula my in textbook wrong? Also for questions the textbook gives me, it sometimes uses di/do and sometimes uses - di/do. IT'S REALLY CONFUSING ME!

Also, can you explain to me all the positive and negatives, that is really confusing, too. Like sometimes numbers will be negative depending on if its a concave mirror or convex or if its a virtual image or what? What's up with that?

Can you explain how to do these? I know the answers, but I'm still confused.
1. A student wants to place an object in front of concave mirror to produce an image half the object's size. If the focal length of the mirror is 5cm, how far form the mirror should the object be places?
This uses the formula M = -di/do when I do it while the next question uses M = di/do.

2. An object place 5cm in front of a concave mirror. The magnification of the object is 2.5. If a real image is produced, what's the radius of curvature of the mirror? (Answer: 16.7)

THANKS SO MUCH!

2. Jul 17, 2010

### ehild

There can be different conventions about the sign of distance of image/magnification. I was taught that the distance of a virtual image from the lens or from the mirror is negative, and so is the magnification.
So the two equations we use are : 1/f = 1/do+1/di and M=hi/ho=di/do. The image size is also negative for a virtual image, but you say the magnitude if the size is the question. The real image is at the reflecting side of a mirror. In case of a lens, the real image is at the opposite side of the lens as the object.
The focal length of a concave mirror is positive, that of a convex one is negative.
In case of thin lenses, the focal length is positive for the converging lenses and negative for the diverging ones.

Using this convention in your first problem, f=5 cm. M=1/2, di/do=1/2, di=do/2,
1/5=1/do+1/di=1/do+2/do ---->1/5=3/do--->do=15 cm.

In the second problem, do=5 cm, a real image is formed, so the magnification is positive. M=2.5, di = 2.5 do= 12.5 cm. The radius of curvature is twice the focal length.
1/f = 1/5+1/12.5--->f=3.57 cm, R= 7.14 cm.