Archived Reflecting telescope calculation

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A reflecting telescope has an objective mirror with a radius of curvature of 3.0m and an eyepiece mirror with a radius of curvature of -1.50m, with a distance of 0.90m between them. The focal lengths calculated are 1.5m for the objective and -0.75m for the eyepiece. The image distance for the objective mirror is determined to be 1.5m, but since the distance to the eyepiece is only 0.9m, the object distance for the eyepiece is -0.6m. Using the lens formula, the final image distance is calculated to be 3m in front of the eyepiece. This indicates where the photographic film should be placed to capture the star's image.
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Homework Statement



A reflecting telescope has a radius of curvature of 3.0m for its objective mirror and a radius of curvature of -1.50m for its eyepiece mirror. If the distance between the two mirrors is 0.90m, how far in front of the eyepiece should you place the photographic film to record the image of a star?

Homework Equations


I could only find out the focal length of both objective mirror and eyepiece mirror.

f=radius of curvature / 2 ...(1)
1/f = 1/di + 1/do ...(2)

The Attempt at a Solution



For now, I got fo=1.5 and fe=-0.75
as the object are from infinity , by using equation 2, di=fo=1.5
Since the distance between the two mirrors is 0.90m , and 1.5-0.9=0.6
0.6 is the object distance of the eyepiece.

1/-0.75 = 1/di + 1/0.6
di is -0.3333
I haven't taught reflecting telescope in my class. I just find a similar solution in yahoo answer.
But I feel the ans is totally wrong. Can anyone help me?
 
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You can always "unfold" a system with mirrors, replacing the mirror with a lens of equal focal length (at least for the purposes of geometric optics). So your approach is basically correct, and I agree that the image distance for the first mirror is 1.5m. But the distance to the second mirror is only 0.9m - so the image (the object for the second mirror) is beyond it. So in the second calculation ##d_o=-0.6##m. Then$$\begin {eqnarray*}\frac 1 {d_i}&=&\frac 1 {0.6}-\frac 1 {0.75}\\
&=&\frac 13\end {eqnarray*} $$and the answer you are looking for is 3m.
 
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