Solving Diverging Mirror Homework: Size & Location of Image

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Homework Help Overview

The problem involves a diverging mirror with a focal length of -60 cm, where a trucker observes the image of a car that is 1.5 m high and located 6.0 m away. The discussion centers around determining the size and location of the image based on the mirror's properties and the sign conventions used in optics.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the image distance (di) using the mirror equation and question the sign conventions applied to object and image distances. There is confusion regarding the negative sign associated with the image distance and the interpretation of the object distance as positive.

Discussion Status

Some participants have confirmed the correctness of certain calculations, while others express confusion about the sign conventions and the implications for the object distance. The discussion reflects a mix of agreement and differing interpretations regarding the textbook's solution and the application of optical principles.

Contextual Notes

Participants note that according to standard sign conventions, object distances should be positive, leading to questions about the textbook's solution and the assumptions made in the problem setup.

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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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