# Refraction of light

1. Feb 6, 2008

### Physicsissuef

Hi! I want to ask you something. When the light travels through medium 1, there is reflection of light, so there is energy released, but it needs time the energy to travel through medium 2, right?

2. Feb 6, 2008

### mgb_phys

Sorry can you try and explain again what you are asking?

3. Feb 6, 2008

### Physicsissuef

Look at this http://www.walter-fendt.de/ph14e/huygenspr.htm" There are 2 mediums? Lets say glass and water. When the light pass through glass it reflects and refracts in same time. It reflects from the glass and refracts in the water, right? Why there is time delay in medium 2?

Last edited by a moderator: Apr 23, 2017 at 10:38 AM
4. Feb 6, 2008

### mgb_phys

when light hits an interface = the boundary between two different refractive indexes some of the light it reflected back (as if from a mirror) and some travels into the second material. This doesn't take any time.
Light does go at different speeds in different materials, the refractive index is (in simple terms) the fraction of the speed of light in vacuum that light travles in that medium.
this doesn't mean any energy is lost.

ps. The applet seems faulty - the wave never gets into the second medium when I run it.

5. Feb 6, 2008

### Physicsissuef

Press 2 times on next step, you'll see that there is time delay.

6. Feb 6, 2008

### Staff: Mentor

When the incoming wave hits one of the pink "atoms" on the boundary, a wave starts speading out immediately from that atom in both directions (upward and downward). The downward-going wave travels more slowly than the upward-going wave, but they both start at the same time.

I think you're seeing what looks like a slight delay in the slower medium because the "atoms" in the diagram are rather large and the waves start from the centers of the atoms. It takes a short time before the wave becomes visible. Before that the pink circle of the "atom" hides the wave.

7. Feb 7, 2008

### Physicsissuef

But look at the angle of refraction, it is different than the angle of reflection. so there is time delay on 2-nd medium

8. Feb 7, 2008

### Andy Resnick

Light scattering can be very non-trivial. The scattering of femtosecond pulses (i.e. the reflection and transmission) is very different than the equilibrium picture we have in textbooks. One must not take cartoon diagrams too seriously.

9. Feb 7, 2008

### Staff: Mentor

Are you really asking, "why is the speed of the wave slower in the second medium than in the first medium?"

I thought you were asking, "why does the (refracted) wave in the second medium start a bit later than the (reflected) wave in the first medium?"

Last edited: Feb 7, 2008
10. Feb 7, 2008

### Physicsissuef

I want to know, is the energy transfer from the atoms of the 1-st medium to the atoms of 2nd medium, reason, of which there is time delay, and the wave of refraction moves slower than the wave of reflection?

11. Feb 7, 2008

### Physicsissuef

yes I could say that my question is: "why does the (refracted) wave in the second medium start a bit later than the (reflected) wave in the first medium?"

12. Feb 7, 2008

### lightarrow

Don't make that animation confuse you: actually reflected and refracted waves starts exactly in the same instant of time, but the lower speed of the refracted wave makes you think it started after a delay. It's not.

13. Feb 7, 2008

### Physicsissuef

Why the refracted wave have lower speed than the reflected wave?

14. Feb 8, 2008

### Physicsissuef

lightarrow, jtbell?

15. Feb 8, 2008

### lightarrow

You have an intuitive answer in the Physics Forums FAQ (one of the first 3 sticky threads of General Physics) with the title "Do Photons Move Slower in a Solid Medium?":

However, solving for Maxwell's equations in a medium, characterized by an electric permittivity and a magnetic permeability

$$\epsilon,\mu$$

you will find that the solution is a function of coordinates and time which propagates with the speed

$$v = 1/\sqrt{\epsilon\mu }$$.

Since, at least for a linear medium, you can write

$$\epsilon\mu = n^2\epsilon_0\mu_0$$

where n is the refraction index > 1, you have:

$$v = 1/n\sqrt{\epsilon_0\mu_0} = c/n < c.$$

Last edited: Feb 8, 2008
16. Feb 8, 2008

### Physicsissuef

Ok, and if I get bigger peace of glass (lets say) and smaller one. Will the index of refraction be same?

17. Feb 8, 2008

### lightarrow

Certainly. It depends only on the chemical composition and on the physical properties of the cristal (at least at non very high frequencies or intensities).

18. Feb 8, 2008

### Physicsissuef

If I get peace of glass and how can I measure its index of reflection? I read the FAQ of the sticky tread, and can't understand still what it want to say, and why it negates the theory of electron transfer of energy.

19. Feb 8, 2008

### lightarrow

I don't know your knowledge of physics, however that is standard physics at first year university and probably even at high school. In the traditional technic you shape the piece of glass as a prism and you measure the vertex angle a and the angle of "least deviation" d_m. Then the refraction index n is: n = [sin(a + d_m)/2]/sin(a/2).
Sincerely I don't know how to explain you better than that way.

20. Feb 8, 2008

### Physicsissuef

Ok, I understand now. Thank you.