Convex Mirrors and Magnification

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The discussion centers on calculating the image distance and magnification for an automobile positioned 4.5 m in front of a convex mirror with a focal length of 0.15 m. The initial calculation for the image distance was found to be 0.155 m, but confusion arose regarding the sign of the focal length, which should be negative for a convex mirror. Participants emphasized the importance of using the correct sign for calculations, leading to clarification on the magnification formula. Ultimately, the original poster resolved the issue, discovering a glitch in the marking program that had marked their answer as correct despite the confusion. The thread concludes with the poster successfully obtaining the correct answers with community assistance.
kc262
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Homework Statement


What is the image distance for an automobile 4.5 m in front of a convex miror with a 0.15 m focal length? I found this to be 0.155 m

The question I'm having a problem with is: What is the magnification?

Homework Equations



magnification = - (image distance/object distance)

The Attempt at a Solution



I thought it would simply be,
-[0.155m/4.5m] = -0.0344. But it's not.

Some help would be much appreciated
 
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kc262 said:
What is the image distance for an automobile 4.5 m in front of a convex miror with a 0.15 m focal length? I found this to be 0.155 m
Careful with signs. For a convex mirror, what sign will the focal length have?
 
Doc Al said:
Careful with signs. For a convex mirror, what sign will the focal length have?

I realized that because it is a convex mirror, the focal length would be negative. However the marking program marked my answer right. I was confused by this :confused: And because of this discrepancy, I am unsure of how to find the magnification.
 
Is the image real or virtual? (And therefore, is the image distance positive or negative?)
 
kc262 said:
I realized that because it is a convex mirror, the focal length would be negative.
Then how did you get an image distance of +0.155 m ? (See Redbelly98's comment.)
However the marking program marked my answer right. I was confused by this :confused:
Now I'm confused. Your first post seemed to indicate that you knew your answer was wrong.
And because of this discrepancy, I am unsure of how to find the magnification.
You find it just as you stated in your first post:
magnification = - (image distance/object distance)
But you need the correct image distance (correct in both magnitude and sign), of course.
 
ok.. i got it. Turns out there was an error/glitch with the marking program. I got the answer in the end. Thanks for everyone's help! :biggrin:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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