# Diverging mirror

1. Jul 2, 2009

### donotremember

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

A trucker sees the image of a car passing her truck in her diverging rear view mirror whose focal length is -60cm. If the car is 1.5m high and 6.0m away, what is the size and location of the image?

2. Relevant equations

1/f = 1/di + 1/do

3. The attempt at a solution

Finding the di

The solution I come up with is:

1/-0.6 = 1/di + 1/6

-1.66 - 0.166 = 1/di

di = -0.5454

The correct answer in my book is :

1/-0.6 = 1/di + 1/-6

-1.66 + 0.166 = 1/di

di = -0.66

Why is the distance to the object negative?

This does not agree with sign convention (as stated in my book)

"Distances of real objects and images are positive"

2. Jul 2, 2009

3. Jul 2, 2009

### donotremember

You mean the distance is not negative and the real answer is -0.5454?

That is strange because the textbook got it wrong and the lecture notes did too.

4. Jul 2, 2009

### rl.bhat

You got di negative because the image is virtual.

5. Jul 2, 2009

### donotremember

By this question i was referring to the 6m distance to the object.

6. Jul 2, 2009

### Redbelly98

Staff Emeritus
This is weird, meaning the book's solution. Object distances are always real (and therefore positive) -- except that when the "object" is actually the image produced by some other mirror or lens, it could be negative. But that's not the case here, the object is an actual object, therefore do must be positive.