Solving Diverging Mirror Homework: Size & Location of Image

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The discussion revolves around calculating the size and location of an image formed by a diverging mirror with a focal length of -60 cm. The initial attempt at solving the problem yielded an image distance (di) of -0.5454 m, while the textbook provided a different answer of -0.66 m. Participants noted that the object distance (do) should be positive, as it refers to a real object, which contradicts the textbook's solution. The confusion stems from the sign convention used in optics, particularly regarding virtual images. Ultimately, the consensus is that the object distance must be positive, affirming the correctness of the initial calculation.
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Homework Statement



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?

Homework Equations



1/f = 1/di + 1/do

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"
 
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Your answer is correct.
 
rl.bhat said:
Your answer is correct.

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.
 
You got di negative because the image is virtual.
 
donotremember said:
You mean the distance is not negative and the real answer is -0.5454?

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