Concave mirror magnification with different object distances

[leolaw]

Each student in a physics lab is assigned to find the location where a bright object may be placed in order that a concave mirror with raidius of curvature r = 40cm will produce an image three times the size of the object. Two students complete the assignment at different times using identical equipment, but when they compare notes later, they discover that their answers for the object distance are not the same. Explain why they do not necessarily need to repeat the lab, and justify you rresponse with a calculation.

Personally, I think that the magnifcation of the object DOES depend on the object distance right?
Because $$\frac{d_0 - f}{f} = \frac{d_0}{d_i}$$

$$d_i (d_0 - f) = f (d_0)$$

$$\frac{d_i}{d_o} = \frac{f}{d_0 - f}$$

$$\frac{-f}{d_0 - f} = m$$ and

$$\frac {-40}{d_0 - 40} = 3$$
which shows that the magnifcation does depend on the object distance.

So why do the question say that they do not have to redo the experiment even if the object distances are different?

 

[learningphysics]

m can take the value -3 or 3. Each gives an image 3 times the size of an object. Hope this helps.

 

[leolaw]

but my point is that, you see $$d_0$$ in the equation? So the magnification does depend on the distance and indeed the object would only magnify three times the size of the original, if it is placed at that location. But the question asks that why you don't have to redo the experiment even though the distances of the object are different from the two student.
So I just don't get the reason of it

 

[leolaw]

hohoho
nvm. you are right -3 and 3 will produce a different $$d_0$$, so if the two students are placing the obejct at exactly those location, then the magnification would be 3, even though one of them may be inverted.

Thx