Magnification for concave mirror

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SUMMARY

A concave mirror produces an image that is 2x larger than the object. By moving both the object and the screen, the image size increases to 3x larger. The screen is moved 0.75 m, prompting the need to calculate the corresponding displacement of the object. The magnification formula, M = - s_i / s_o, is essential for deriving the relationship between the distances of the image and object.

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  • Understanding of concave mirror properties
  • Familiarity with the magnification formula (M = - s_i / s_o)
  • Basic algebra for solving equations
  • Knowledge of image formation in optics
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jaejoon89
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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!
 
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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.
 

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