- #1
Roger Dodger
- 42
- 3
I have a conundrum of sorts that has made me feel like an idiot and I am hoping someone can point out my mistake.
Suppose a light source is placed to the left of a prism and a detector is placed on the opposite side. I have seen plenty of pictures of this sort, and they all appear to show the light passing through the prism obeying Snell's Law. But in doing so, they have the blue light sampling a more substantial portion of the prism than the red light. To me, this makes no sense. Blue light is slower than red light when passing through glass so if anything it would want to sample those paths that reduce its time in the prism.
I fail to see how Snell's Law applies here. Every derivation I have seen of Snell's Law has the detector placed inside the glass. This will give a local path of least action for light refracting from the air into the glass. However, this path would not appear to be related to the path of least action for the entire trip from source to the detector placed to the right of the prism.
Using Feynman's QED description of light, the probability amplitude for reaching the detector would be the product of (1) the probability amplitude for reaching a detector inside the prism and (2) the probability amplitude for light emanating from that detector into the air and reaching the other detector.
I realize we can see the light as it passes through the detector, but that's a completely different problem, since there is no guarantee that the light that reaches the detector sampled the same paths that the light that reaches our eyes sampled.
Or do I have it all wrong?
Suppose a light source is placed to the left of a prism and a detector is placed on the opposite side. I have seen plenty of pictures of this sort, and they all appear to show the light passing through the prism obeying Snell's Law. But in doing so, they have the blue light sampling a more substantial portion of the prism than the red light. To me, this makes no sense. Blue light is slower than red light when passing through glass so if anything it would want to sample those paths that reduce its time in the prism.
I fail to see how Snell's Law applies here. Every derivation I have seen of Snell's Law has the detector placed inside the glass. This will give a local path of least action for light refracting from the air into the glass. However, this path would not appear to be related to the path of least action for the entire trip from source to the detector placed to the right of the prism.
Using Feynman's QED description of light, the probability amplitude for reaching the detector would be the product of (1) the probability amplitude for reaching a detector inside the prism and (2) the probability amplitude for light emanating from that detector into the air and reaching the other detector.
I realize we can see the light as it passes through the detector, but that's a completely different problem, since there is no guarantee that the light that reaches the detector sampled the same paths that the light that reaches our eyes sampled.
Or do I have it all wrong?