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I barely know anything about optics, so I could use some help about how to go about solving this problem.

If I have a ray of light at a certain height from the optical axis, propagating at an angle, and the distance it travels before it comes in comes in contact with a mirror, can I find the angle and height of the ray after it reflects?

I know the mirror is concave and the radius of curvature, and I know the parameters listed above, but how should I write an equation to find the height and angle of a reflected ray?

EDIT: For example, I have a ray 5mm from the optical axis, and it propagates at 0.3 mrad to that axis. The ray travels a distance of 2cm and comes in contact with a concave mirror of R = 1. How can I find the height and angle of the ray 1m away from the point reflected? I've looked at three optics books, and I still can't find an easy explanation for calculating this.

I tried to use the matrix method to solve this problem, but I don't even know if this would work or not. Below is what I referred to. Can anyone tell me if this would work for my problem?

Concave reflection

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2 Answers

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calculation is easy you can apply the mirror formula and magnification formula which will give yo the position of the new image and it's magnification through which you can find the angle of the reflected ray. If the travelling ray is striking at the pole of the mirror then the the reflected ray will be of same length and height from the optical axis and you will get the height of the object. But it is applicable only for point object. If it is a elongated object of height h from the optical axis then you will have to take two rays ray 1 and 2 as given in the fig. The point of intersection of the two ray's will give you the distance of the object from the pole p. Two find the distance x apply the following formula (1/u + 1/v = 1/f) where u is the distance of the object from p v is the distance of the image from the p and f is the focal length. And for height apply magnification formula (-v/u = h'/h) where h' is height of image and h is height object. So I hope it may help you. (Remember to use sign convention. Keep following points in your mind. Al the distance are measured from pole in either direction's. Distance measured from the pole which are in the direction of incident ray are positive and in th opposite direction are negate si while placing the value of u and v apply the sign convention) this is the image

enter image description here

Her is the url of the animation hope it will help you more http://www.physicsclassroom.com/mmedia/optics/rdcma.cfm

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I don't understand. I'm not talking about images. I have a ray, and it reflects off the mirror and I want to find where it goes. The ray is not parallel to the optical axis. – Chris Harris Mar 2 at 3:23
Yes, if you will calculate the position of the image then you can calculate the angle towards which the incident ray from the object will get reflected. Any ray which will strike the mirror below the ray 2 it will get reflected towars the point wherw the image will be formed so in optics we only take two ray's one which is parallel to the principal axis and other which passes through p or focus. I think you are not familiar to optic's at all in which class do you study if you don't mind. – Akash Mar 2 at 3:54
I'm taking a course in lasers, and you are right, I'm not familiar with optics. I don't understand because the ray is not incident with the pole of the mirror. It is not parallel either. – Chris Harris Mar 2 at 4:05
Don't worry i told you any ray which get reflected from the mirror from any point goes towards the point where image will be finally formed. I have uploded one more image to make you understand. An applying simple geometrical calculation's you can calculate it. This is the answer if interpreted your's question correctly. If you think I am understanding the ques in a different way than you want the you can correct me. An if i am right and you still don't understand then you can refer to the optic's book of class 12th. – Akash Mar 2 at 4:17
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The ray will reflect at the negative of the incident angle with respect to the normal, so all you have to do is compute the normal (which you can obtain from the derivative of the surface curve), then find the perpendicular vector to that normal. That is it...

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