Location of a virtual image in a convexed mirror

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  • #1
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Homework Statement


An object 2.71 * 10-2m tall is 0.14m in front of a convex mirror that has a radius of curvature 42cm. Calculate the position and height of the image, is the image inverted or upright?


Homework Equations


Well, I've stated that I'm using the ray model of light.
As far as I can deduce, the equation Focal length = R/2, 1/u + 1/v = 1/f.
magnification = v/u.
Where u = distance from mirror to object
v = distance from mirror to image
f = focal length.
r = center of curvature


The Attempt at a Solution


I have found the focal length to be 0.21m, and the image distance being 0.07m by rearranging 1/u + 1/v = 1/f.
I have found the height of the image to be half the original height.
I'm not sure if this is right, as I feel I have made up how to go about it, any pointers?
 

Answers and Replies

  • #2
PeterO
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Homework Statement


An object 2.71 * 10-2m tall is 0.14m in front of a convex mirror that has a radius of curvature 42cm. Calculate the position and height of the image, is the image inverted or upright?


Homework Equations


Well, I've stated that I'm using the ray model of light.
As far as I can deduce, the equation Focal length = R/2, 1/u + 1/v = 1/f.
magnification = v/u.
Where u = distance from mirror to object
v = distance from mirror to image
f = focal length.
r = center of curvature


The Attempt at a Solution


I have found the focal length to be 0.21m, and the image distance being 0.07m by rearranging 1/u + 1/v = 1/f.
I have found the height of the image to be half the original height.
I'm not sure if this is right, as I feel I have made up how to go about it, any pointers?
Don't forget that for a convex lens, the focal length is negative.

EDIT: and best of all - draw a ray diagram!!!
 
Last edited:
  • #3
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That was one of my problems.
I drew a ray diagram, and it places the image between the mirror and the focal length, so I assume the image was located at <21cm. But re-arranging the equation, if I keep the distance for f negative, I receive an answer that is not in agreement with the ray diagram.
 
  • #4
PeterO
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That was one of my problems.
I drew a ray diagram, and it places the image between the mirror and the focal length, so I assume the image was located at <21cm. But re-arranging the equation, if I keep the distance for f negative, I receive an answer that is not in agreement with the ray diagram.
I think you also get image distance as negative meaning the image, like the focus, is behind the mirror.
 
  • #5
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Ahh! I see. I think I must have messed up my ray diagram, because I think my maths is solid.
 
  • #6
PeterO
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Ahh! I see. I think I must have messed up my ray diagram, because I think my maths is solid.
I got a little over 0.08m so if your maths gives you 0.07 I would be worried.

Once you get the right position, you will get the right height [magnification] and whether upright / inverted.
 
  • #7
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Sorry if I'm being a pain, but is this how you got it?

R = 0.42.
f = -0.42/2 = -0.21
(1/.14)+(1/v) = (1/-.21)
(1/v) = (1/-.21)-(1/.14)
1/v = -11.905
1/-11.905 = v = -.084
?
 
  • #8
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Of course, if its against the forums rules to say yes or no. My apologies.
I haven't read them in a while.
 
  • #9
PeterO
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Sorry if I'm being a pain, but is this how you got it?

R = 0.42.
f = -0.42/2 = -0.21
(1/.14)+(1/v) = (1/-.21)
(1/v) = (1/-.21)-(1/.14)
1/v = -11.905
1/-11.905 = v = -.084
?
EXACTLY. Now for the size of the image and whether it is upright or inverted.
 
  • #10
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The magnification formula is m = v/u
v = -0.084 u= 0.14
thus m = -.6
This implies that the image is inverted.
However, my ray diagram begs to differ. I think I need a little more practices with ray diagrams.

Edit:
I found a source quoting that the m = -v/u
If this is true, then all my problems are solved I think.
 
  • #11
PeterO
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The magnification formula is m = v/u
v = -0.084 u= 0.14
thus m = -.6
This implies that the image is inverted.
However, my ray diagram begs to differ. I think I need a little more practices with ray diagrams.

Edit:
I found a source quoting that the m = -v/u
If this is true, then all my problems are solved I think.
Calculation is perfect, interpretation failed.

A negative magnification means upright.
 
  • #12
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Oh, Ok. Thanks.
Although, this means the site I was given for this was wrong. It clearly states the opposite.
However, it does make more sense this way.
Thank you kindly.
 
  • #13
PeterO
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Oh, Ok. Thanks.
Although, this means the site I was given for this was wrong. It clearly states the opposite.
However, it does make more sense this way.
Thank you kindly.
Interesting concept.
I don't throw a - sign in the formula, but say + is inverted, - is upright.

Your reference [and wikipedia I noticed] have a - sign in their formula so that + is upright and - is inverted.

I must say I use ray tracing and a different formula altogether to calculate all lenses and mirrors anyway. My formula is a mathematical summary of the ray trace, was the common formula in texts in the 1960s, and has the square of f included, so you could never make the mistake of not recognising a negative focal length.
 
  • #15
PeterO
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Ahh. I see, thanks.

If it interests you, my two conflicting sources, which must have caused me some confusion were

http://www.physics.uq.edu.au/people/mcintyre/vergences/optics/go_mag.html
http://www.physicsclassroom.com/class/refln/u13l4d.cfm

Thanks once more!
The sign system I use, is that + equates to the normal or most common.

If you use a convex lens - a common magnifying glass - then of all the positions you can place an object, a small distance from the lens can be described as "inside the focus" while the rest of infinity is "outside the focus". That tells me that an object is more likely to be found outside the focus than inside the focus.
When an object is placed outside the focus, the image is inverted - and behind the lens: able to be cast on a screen. To me then all those properties are positive.

So a positive result means:
On the other side of the lens, inverted, able to be cast on a screen [real]

I carry that same sign convention over to concave lenses, and both sorts of mirrors.
 
  • #16
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So, just say that the above question was for a concave mirror instead of a convexed mirror, how would the equations differ?
 
  • #17
PeterO
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So, just say that the above question was for a concave mirror instead of a convexed mirror, how would the equations differ?
The equations would not differ, but the focus would be positive.

Qualitativly: an object 14 cm from a concave mirror with 21 cm focal length means the image is: behind the mirror [Di would be negative]
The magnification would be negative, meaning upright image, unable to be cast on a screen - so virtual. from my formula I know would be 42 cm behind the mirror.

F = 21, Object is 7 cm from the focus 21*21 = 7 x So ---> Image is 63 cm from focus so 42 behind the mirror. The image is 3 times the size of the object.
 

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