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Biker
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So we are studying optics in school this semster, Very interseting topic I say but I just have a couple of question I want to ask.
In concave and convex mirror, we study spherical ones where F = R/2. I was able to prove this and that it is only an approximation when ## R >> h_o ## or ## h_0## is small enough and as concave mirror and convex mirror are the same this applies to both. But in lenses I couldn't do that, I read in my book that they assume that the lens is thin and it only refract once in the center of the lens. I don't really understand that.
What should happen is that first I have to measure the derivative of the spherical mirror at the point where the light hits the surface of the mirror then take the perpendicular and use snell's law, The light travels a bit inside the lens so it hits a different point on the other surface of the mirror so I need to do what I did once again.
So What I am asking is how does the thin lens approximation make it refract at a single point? and a Text book that I have said that the focal point of lens is R/2 which is wrong I guess and Is there a derivation which consider lens purely mathematical to compute the focal length? without assumptions(Only thin lens)?The other thing is about apparent depth, Let's first assume an object which it's depth is ##y_o## and you make a perpendicular ray coming out of it and let's assume that the eye of the observer is near the perpendicular line.
Since our eye is not that big that eye will only see two rays which the distance between them on the surface is small (Or you can say the angle between them is small whatever you like)
So the 2nd ray will intersect with the perpendicular one at Point ## Y_i ## and it can be calculated through the equation I derived:
##(\frac{dx^2+ y_i^2}{dx^2 + y_o^2})^{0.5} = 1/n##
Where this object is under for example water and the ray is going to air.
Now as dx gets really small the approximation of ## n = \frac{y_o}{y_i} ## gets better. I guess even if there is difference in Yi the brain enhances it to appear coming of 1 point. All of this to answer one question what happens when you are exactly at the perpendicular? I read a textbook that says that you see the actual depth since light doesn't refract. But the problem is the eye isn't a point, it can never know where the object is if you only have one ray.. However, If you take dx as infinitesimal you get the same equation above to approximate the situation or you can say that the center of the eye is on the object but since it has width and that it is small you get the same answer again.
So If I have an exam, What answer should I consider?
In concave and convex mirror, we study spherical ones where F = R/2. I was able to prove this and that it is only an approximation when ## R >> h_o ## or ## h_0## is small enough and as concave mirror and convex mirror are the same this applies to both. But in lenses I couldn't do that, I read in my book that they assume that the lens is thin and it only refract once in the center of the lens. I don't really understand that.
What should happen is that first I have to measure the derivative of the spherical mirror at the point where the light hits the surface of the mirror then take the perpendicular and use snell's law, The light travels a bit inside the lens so it hits a different point on the other surface of the mirror so I need to do what I did once again.
So What I am asking is how does the thin lens approximation make it refract at a single point? and a Text book that I have said that the focal point of lens is R/2 which is wrong I guess and Is there a derivation which consider lens purely mathematical to compute the focal length? without assumptions(Only thin lens)?The other thing is about apparent depth, Let's first assume an object which it's depth is ##y_o## and you make a perpendicular ray coming out of it and let's assume that the eye of the observer is near the perpendicular line.
Since our eye is not that big that eye will only see two rays which the distance between them on the surface is small (Or you can say the angle between them is small whatever you like)
So the 2nd ray will intersect with the perpendicular one at Point ## Y_i ## and it can be calculated through the equation I derived:
##(\frac{dx^2+ y_i^2}{dx^2 + y_o^2})^{0.5} = 1/n##
Where this object is under for example water and the ray is going to air.
Now as dx gets really small the approximation of ## n = \frac{y_o}{y_i} ## gets better. I guess even if there is difference in Yi the brain enhances it to appear coming of 1 point. All of this to answer one question what happens when you are exactly at the perpendicular? I read a textbook that says that you see the actual depth since light doesn't refract. But the problem is the eye isn't a point, it can never know where the object is if you only have one ray.. However, If you take dx as infinitesimal you get the same equation above to approximate the situation or you can say that the center of the eye is on the object but since it has width and that it is small you get the same answer again.
So If I have an exam, What answer should I consider?