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Optics (refracting light) and (mirrors)

  • Thread starter Jehuty
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1)For my homework, I received a question, I'm supposed to use Snells law and a protractor. Basically, I have an odd shaped object that forms a messed up sphere. I have an arrow going in at a specific point and I'm given

n = 1.0 (for the light ray outside the sphere) and
n = 1.2 (for inside the sphere)

I know how to apply the two numbers into snells law but I don't know how I'm supposed to come up with an angle. It is not a triangle so I can't use trigenometry to figure it out.

2) Proove mathematically using the mirror equation that when an object is far away from a concave mirror, its image can be found at the focal point. Explain any assumptions.

As far as I got with question 2 is di = f I simply can't understand this.
 

berkeman

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-1- You need to draw a line that is tangent to the lens/sphere surface at the point where your ray is entering. The normal to that tangent is the normal that you use with Snell's law. For a lens, the normal is is at a different angle usually for each ray that you trace through the lens.

-2- What is the mirror equation. What value of distance do you put in for the object distance if it is "very far away"?
 

Doc Al

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Jehuty said:
1)....I know how to apply the two numbers into snells law but I don't know how I'm supposed to come up with an angle.
Measure your angles with respect to the normal to the surface. (The normal is perpendicular to the surface.)

2) ...As far as I got with question 2 is di = f I simply can't understand this.
It looks like you've got it to me. Apparently you've shown that the image is a distance f from the center of the mirror, which is the location of the focal point. (You may need to refresh your understanding of the mirror equation and what it means.)
 
14
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Thank you for all your help with question 1, I'm sorry I didn't state question 2 clearly. The mirror equation I am using is H(image) / H(object) = D(image) / D(object) where H is height and D is distance. With question 2, I know vaguely of what I am supposed to get but I am having a hard time mathematically getting the answer. I'm assuming that "very far away" means anything past the focal point and the curvature.
 

berkeman

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Is that the mirror equation that they are referring to? Remember, it's a concave mirror. I'd expect more like a lens-type equation....
 
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Lens equation....isn't the lens equation the same except it has a negative for the distance image?
 

Doc Al

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Jehuty said:
The mirror equation I am using is H(image) / H(object) = D(image) / D(object) where H is height and D is distance.
That's a description of linear magnification. What's usually called the "mirror equation" (similar to the thin spherical lens equation) is something like this:
[tex]\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}[/tex]
 

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