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Really simple question about mirrors
Where must you place an object in front of a concave mirror with radius R so that the image is erect and 2.5 times the size of the object? Where is the image?
[tex]1/s + 1/s' = 2/R[/tex]
[tex]m = h'/h = -s'/s[/tex]
I know that the object will be within the focal length of the mirror, and the image will be virtual. However, I'm having problems with the signs of my values. I know that the object distance will be positive, the radius of curvature will be positive, and the image distance will be negative. What's the right way to put this into the equations?
From [tex]m = h'/h = -s'/s[/tex] I get [tex]s = -s'/2.5[/tex], and from there, I substitute s into [tex]1/s + 1/s' = 2/R[/tex] to get [tex] 1/s - 1/(s'/2.5) = 2/R[/tex]. This gives me [tex] s' = -3/4 R[/tex]. I don't think this is right, as I get a negative object distance and positive image distance, which shouldn't happen. Can anyone give me some help with this?
Homework Statement
Where must you place an object in front of a concave mirror with radius R so that the image is erect and 2.5 times the size of the object? Where is the image?
Homework Equations
[tex]1/s + 1/s' = 2/R[/tex]
[tex]m = h'/h = -s'/s[/tex]
The Attempt at a Solution
I know that the object will be within the focal length of the mirror, and the image will be virtual. However, I'm having problems with the signs of my values. I know that the object distance will be positive, the radius of curvature will be positive, and the image distance will be negative. What's the right way to put this into the equations?
From [tex]m = h'/h = -s'/s[/tex] I get [tex]s = -s'/2.5[/tex], and from there, I substitute s into [tex]1/s + 1/s' = 2/R[/tex] to get [tex] 1/s - 1/(s'/2.5) = 2/R[/tex]. This gives me [tex] s' = -3/4 R[/tex]. I don't think this is right, as I get a negative object distance and positive image distance, which shouldn't happen. Can anyone give me some help with this?
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