How Does Snell's Law Explain Light's Unchanged Direction Through a Glass Plate?

AI Thread Summary
Snell's Law explains that when a light ray passes through a glass plate, it refracts at both the air-glass interface and the glass-air interface. The formula sin(theta in)/sin(theta refracted) = n2/n1 is applied to demonstrate that the angle of incidence and angle of refraction balance out. The light ray bends toward the normal when entering the glass and bends away from the normal when exiting, resulting in no net change in direction. This phenomenon occurs because the refractive indices of air and glass differ, but the double refraction compensates for each other. A clear understanding of this process can be enhanced by visualizing it with a diagram.
clipperdude21
Messages
48
Reaction score
0
Snell's Law and Light!

1. Use Snell's Law to show that a light ray which enters glass plane of wfinite width does not change direction when it exits the plate.



2. sin(theta in)/sin(theta refracted)= n2/n1



3. I know that if lamda in medium 1 does not equal lamda in medium 2, the direction of the wave changes. To show that the wave doesn't change directions we would need to show that lamba 1= lamda 2?
 
Physics news on Phys.org
You need to think about this a little more. There is no way that the index of refraction for air will ever be equal to the index of refraction of glass. What you are missing is a clear picture of what is happening in this set up. It helps to draw a diagram.

The light goes from medium 1 (air) into medium 2 (glass) and then BACK into medium 1 (air). In other words, it goes THROUGH the glass plate. That means that it gets refracted TWICE. Once at the air-glass interface on the near side of the plate, and once at the glass-air interface on the FAR side. So you have to use Snell's law twice. Once for the first refraction, and once for the second refraction, in order to show that the amount by which the beam is bent toward the normal upon entering the glass is exactly compensated for by the amount by which it is bent away from the normal upon exiting. Then you will have shown that there is no NET refraction.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top