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I just had a bit of an eureka moment, so I thought I would share it with SDN in case it helps any of you guys deal with optics problems. Feel free to correct me if anything is wrong here, since it's rather late (2:30 AM). Also, sorry if it's a bit hard to understand since I typed it up quickly at 2:30 AM lol.
Basically, you don't need to memorize ray diagrams, when images will be virtual/real, when images will be reduced/enlarged, etc.
You only have to know the following two formulas. With some basic math, you can figure out the rest.
1/i+1/o=1/f
magnification=-1/o
And know these basic conditions:
Converging mirror/lens: f>0
Diverging mirror/lens: f<0
For any single mirror/lens system, o (object distance) will always be greater than 0 since the object is being placed in front of the lens or mirror.
With that said, here's how you can apply the 2 formulas and the bits of basic knowledge above to solve optics problems without rote memorization:
Converging mirror/lens:
f>0. Since o is always >0, and 1/o+1/i=1/f, we have the following two possibilities for i:
1) When o>f (object placed outside focal length), 1/o is smaller than 1/f, and so i must be positive for the equation to hold.
When i is positive, magnification is negative, and thus you have a real, inverted image.
2) When o<f (object placed inside focal length), i must be negative for the equation to hold.
When i is negative, magnification is positive, and thus you have a virtual, upright image. In addition, |1/i| must be smaller than |1/o| (or else you would get a negative value for 1/f). thus, i must be greater than o, and the image will be enlarged.
Diverging mirror/lens:
f<0 and o is always >0
In order for 1/o+1/i=1/f to be true, i must always be <0. Thus, magnification is positive and you get an upright, virtual image. In addition, since 1/f is negative, |1/i| must be larger than |1/o|, so i must be smaller than o and the image will be reduced.
Basically, you don't need to memorize ray diagrams, when images will be virtual/real, when images will be reduced/enlarged, etc.
You only have to know the following two formulas. With some basic math, you can figure out the rest.
1/i+1/o=1/f
magnification=-1/o
And know these basic conditions:
Converging mirror/lens: f>0
Diverging mirror/lens: f<0
For any single mirror/lens system, o (object distance) will always be greater than 0 since the object is being placed in front of the lens or mirror.
With that said, here's how you can apply the 2 formulas and the bits of basic knowledge above to solve optics problems without rote memorization:
Converging mirror/lens:
f>0. Since o is always >0, and 1/o+1/i=1/f, we have the following two possibilities for i:
1) When o>f (object placed outside focal length), 1/o is smaller than 1/f, and so i must be positive for the equation to hold.
When i is positive, magnification is negative, and thus you have a real, inverted image.
2) When o<f (object placed inside focal length), i must be negative for the equation to hold.
When i is negative, magnification is positive, and thus you have a virtual, upright image. In addition, |1/i| must be smaller than |1/o| (or else you would get a negative value for 1/f). thus, i must be greater than o, and the image will be enlarged.
Diverging mirror/lens:
f<0 and o is always >0
In order for 1/o+1/i=1/f to be true, i must always be <0. Thus, magnification is positive and you get an upright, virtual image. In addition, since 1/f is negative, |1/i| must be larger than |1/o|, so i must be smaller than o and the image will be reduced.
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