Solving Diverging Mirror Homework: Size & Location of Image

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In summary, the conversation discusses the calculation of the size and location of an image seen in a trucker's diverging rear view mirror. The focal length is -60cm and the car passing the truck is 1.5m high and 6.0m away. The conversation also mentions the use of the equation 1/f = 1/di + 1/do and the correct answer being di = -0.66. The confusion arises when the distance to the object is negative, which goes against the sign convention stated in the textbook. The expert confirms that the distance should not be negative in this case and the real answer is actually -0.5454. The book's solution is incorrect as it considers the image to
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donotremember
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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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  • #2
Your answer is correct.
 
  • #3
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.
 
  • #4
You got di negative because the image is virtual.
 
  • #5
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.
 
  • #6
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.
 

FAQ: Solving Diverging Mirror Homework: Size & Location of Image

1. How do I determine the size of the image in a diverging mirror?

The size of the image in a diverging mirror can be determined by using the magnification formula, M = -di/do, where M is the magnification, di is the distance from the mirror to the image, and do is the distance from the mirror to the object. A negative magnification indicates a virtual image that is upright and smaller than the object.

2. What factors affect the location of the image in a diverging mirror?

The location of the image in a diverging mirror is affected by the distance of the object from the mirror, the focal length of the mirror, and the angle at which the light rays hit the mirror. These factors can be used to calculate the distance from the mirror to the image using the mirror equation, 1/f = 1/di + 1/do, where f is the focal length of the mirror.

3. How does the location of the object affect the size of the image in a diverging mirror?

The size of the image in a diverging mirror is directly proportional to the distance of the object from the mirror. As the object gets closer to the mirror, the image will appear larger. This can be seen in the magnification formula, where a smaller do will result in a larger image.

4. Can the size of the image in a diverging mirror be larger than the object?

No, the image in a diverging mirror can never be larger than the object. This is because the light rays diverge away from each other, causing the image to always be smaller than the object. The magnification for a diverging mirror is always less than 1, indicating a reduction in size.

5. How do I know if the image formed by a diverging mirror is real or virtual?

The location of the image in a diverging mirror can be used to determine if it is real or virtual. A virtual image will always be located behind the mirror, while a real image will be located in front of the mirror. Additionally, a virtual image will always be upright, while a real image may be inverted depending on the location of the object.

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