Convex Mirrors and Magnification

AI Thread Summary
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
Messages
4
Reaction score
0

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
 
Last edited:
Physics news on Phys.org
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:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top