magnification equation

[Addallat]

Two part question:


1. I just need someone to confirm the following for me:

when the absolute value of m < 1 then the image is reduce in size

when the absolute value of m > 1 the image is enlarged in size

the fact that m is positive just tells you that i is negative and that the image if upright and virtual (vice versa for negative with IR) and the sign of m has nothing to do with whether the image is enlarged or shrunken, right?


2. for the magnification equation

m = - (i)/(o)

is i and o:
the distances of the image and object
or
the height of the image and the object
?


I'm really confused on my second question; I just did a problem that gave me the distance of the object and the radius of curvature which i used the mirror equation to solve for the distance of the image (using focal length = 1/2r), but then in the second part of the question they took the values for the distance of the image and the distance of the object and plugged them into the magnification equation!?!?!

So I'm guessing you can use heights or distances for the magnification equation???

TPR really does a horrible job on explaining optics they completely neglected to mention how to interpret whether image is enlarged or reduced in size and state nothing about object distance and object height being permissible for the magnification equation

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[Bomikepa]

1. I just need someone to confirm the following for me:

when the absolute value of m < 1 then the image is reduce in size

Yes- any value lower than 1 is smaller image

when the absolute value of M > 1 the image is enlarged in size.

yep

the fact that m is positive just tells you that i is negative and that the image if upright and virtual (vice versa for negative with IR) and the sign of m has nothing to do with whether the image is enlarged or shrunken, right?

yes the sign does not tell you anything about the relative size of the image

the sign of the Magnification equation tell you if the image is upright or inverted, and based on this you can determine if the image is real or virtual

real goes with inverted
virtual goes w upright


2. for the magnification equation

m = - (i)/(o)

is i and o:
the distances of the image and object
or
the height of the image and the object


You are very close but there are two equations for magnification. One involves distances and the other involved heights.

M=hi/ho=- di/do

hi=height image
ho=height object
and the d is the distance image and object

hope this clears something up

 

[Addallat]

You are very close but there are two equations for magnification. One involves distances and the other involved heights.

M=hi/ho=- di/do

hi=height image
ho=height object
and the d is the distance image and object

hope this clears something up

oh man, so wait for there's no negative sign for hi/ho???

m = hi/ho



m = - (di)/(do)



Please confirm that the above two are correct, thank you!

Edit: this site states that there's no negative sign for the hi/ho magnification equation
http://www.physicsclassroom.com/Class/refrn/u14l5f.cfm ; again just wanted to point out TPR did such a sucky job on optics (skip chap 11 tpr and use another source imo) i get the fact that height can't be negative but still they neglect to mention you can't use heights OR distances to solve for magnfication

 

[Bomikepa]

oh man, so wait for there's no negative sign for hi/ho???

m = hi/ho



m = - (di)/(do)



Please confirm that the above two are correct, thank you!

yes- NO NEGATIVE for the height equation for magnification.

 

[Bomikepa]

post the question and il try and solve it.

yes the height equation has no negative only the distance one has a negative to account for the negative images.

M=hi/ho M= - di/do maybe just keep them separate to avoid confusion