Geometric Optics - Magnification

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SUMMARY

The forum discussion centers on a geometric optics problem involving a concave mirror, where an image is initially twice the size of the object and later becomes three times larger after both the object and screen are moved. The key equations discussed include the magnification formula, m = image distance / object distance, and the mirror equation, 1/d_I + 1/d_o = 1/f. The participants emphasize the need to establish relationships between the object and image distances to solve for how far the object moved, given that the screen was moved 75 cm.

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LANS
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Homework Statement


A concave mirror forms an image on a screen twice as large as an object. Both object and mirror are then moved such that the new image is 3x the size of the object. If the screen is moved 75cm, how far did the object move?

Homework Equations


m = image distance / object distance

The Attempt at a Solution


Initially, the screen is twice as far from the mirror than the object. The screen and object move so that the screen is three times as far from the mirror as the object. I'm really at a loss and feeling kinda dumb here on solving this - any pointers in the right direction would be much appreciated.

Thank you,
LANS
 
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Is that exactly how the question was worded? It sounds a bit off because it says the object and mirror are moved, implying the screen stays put, but then it says the screen was moved.
 
vela said:
Is that exactly how the question was worded? It sounds a bit off because it says the object and mirror are moved, implying the screen stays put, but then it says the screen was moved.

Sorry, I miswrote it. The object and screen moved, mirror stays in the same place.
 
LANS said:

Homework Statement


A concave mirror forms an image on a screen twice as large as an object. Both object and mirror are then moved such that the new image is 3x the size of the object. If the screen is moved 75cm, how far did the object move?

Homework Equations


m = image distance / object distance

The Attempt at a Solution


Initially, the screen is twice as far from the mirror than the object. The screen and object move so that the screen is three times as far from the mirror as the object. I'm really at a loss and feeling kinda dumb here on solving this - any pointers in the right direction would be much appreciated.
OK, let's define some variables:
\begin{eqnarray*}
d_\text{obj} &= \text{initial object distance} \\
d_\text{img} &= \text{initial image distance} \\
r_\text{obj} &= \text{final object distance} \\
r_\text{img} &= \text{final image distance} \\
f &= \text{focal length of the mirror}
\end{eqnarray*} So you have five unknowns. You have ##d_\text{img}/d_\text{obj} = 2## and ##r_\text{img}/r_\text{obj} = 3## so far. That's two equations. Can you come up with any more?
 
There's 1/d_I + 1/d_o = 1/f (and same for r_I, r_o), and r_I - d_I = 75, I'm just not sure how to approach it algebraically.
 
OK, good. Since you're looking for how far the object moved, you might want to get equations in terms of the object distances, so I'd use the magnification equations to solve for the image distances in terms of the object distances and then substitute the results into the three remaining equations.

Try writing those out and see if you can see where to go from there. Sometimes you just have to write equations down and try stuff, and then it becomes clear what to do.
 

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