Magnification for concave mirror

[jaejoon89]

A concave mirror creates an image on the screen 2x larger than the object. BOTH objects and screen are subsequently moved in order to create an image on the screen 3x larger than the object. If the screen is moved 0.75 m in this process, how far is the object also moved?

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Magnification, M = - s_i / s_o
where s_i and s_o are the distances of the image and object, respectively

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I think the wording's throwing me off. It says the screen is moved--so does that mean the position s_i2 isn't actually 0.75 m but this number is just the displacement from the first value?

In which case:
2 = - s_i1 / s_o1
3 = - s_i2 / s_o2 = - (s_i1 +/- 0.75m) / s_o2 (I'm not sure if it should be + or -)
This would be 2 equations, 3 unknowns so I must be doing something wrong. Please help!

 

[Carid]

Yes, it's a displacement value.
You can derive an expression for the object displacement in terms of the original object distance. I can't see anything else.

 

[nlightner]

I would approach this problem by first clarifying the language used. You are correct in questioning whether the 0.75 m is the actual displacement or just the distance traveled. This information is important in determining the appropriate equations to use.

Assuming that the 0.75 m is the actual displacement of the screen, we can use the following equations:

M1 = - s_i1 / s_o1
M2 = - s_i2 / s_o2 = - (s_i1 +/- 0.75m) / s_o2

We know that M1 = 2 and M2 = 3, so we can set up the following equations:

2 = - s_i1 / s_o1
3 = - (s_i1 +/- 0.75m) / s_o2

Solving for s_o1, we get s_o1 = -s_i1/2. Substituting this into the second equation, we get:

3 = - (s_i1 +/- 0.75m) / (-s_i1/2)
3 = (s_i1 +/- 0.75m) / (s_i1/2)
6 = s_i1 +/- 0.75m
s_i1 = 6.75m or 5.25m

This means that the object must be moved either 6.75 m or 5.25 m, depending on whether the screen was moved towards or away from the object. This makes sense because moving the object closer or farther from the mirror would affect the magnification.

In conclusion, as a scientist, I would recommend clarifying the language used in the problem to ensure the correct equations are used. The final answer would depend on whether the 0.75 m is the actual displacement of the screen or just the distance traveled.