Geometric Optics - Magnification

In summary: Remember that the goal is to get an equation that relates ##r_\text{obj}## to ##d_\text{obj}##.In summary, a concave mirror forms an image on a screen twice as large as the object. When both the object and mirror are moved, the new image is 3 times the size of the object. Using the equations for magnification and focal length, we can write a system of equations to solve for the distance of the object. By substituting the results of the magnification equations into the equations for focal length and distance moved, we can ultimately find an equation that relates the initial and final distances of the object. This can then be solved to determine how far the object moved.
  • #1
LANS
24
0

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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  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
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.
 
  • #6
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.
 

1. What is magnification in geometric optics?

Magnification in geometric optics is a measure of how much an object appears to be enlarged or reduced when viewed through an optical instrument, such as a lens or mirror. It is calculated by dividing the size of the image by the size of the object.

2. How is magnification related to the focal length of a lens?

The magnification of a lens is inversely proportional to its focal length. This means that as the focal length increases, the magnification decreases, and vice versa. This relationship is described by the magnification formula: M = -di/do, where M is the magnification, di is the image distance, and do is the object distance.

3. Can magnification be negative?

Yes, magnification can be negative when the image formed by an optical instrument is inverted, meaning it appears upside down compared to the object. Negative magnification is often used to indicate that the image is smaller than the object.

4. How does magnification affect the size and clarity of an image?

The higher the magnification, the larger the image appears. However, as magnification increases, the image also becomes more distorted and less clear. This is due to factors such as aberrations and diffraction, which can affect the quality of the image.

5. Is there a limit to how much magnification can be achieved in geometric optics?

Yes, there is a limit to the amount of magnification that can be achieved in geometric optics. This is known as the optical resolution limit, and it is determined by the numerical aperture of the optical system. Beyond this limit, the image will become increasingly blurry and distorted.

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