# Reflection and Wavefronts

1. Sep 4, 2011

### Peter G.

Hi,

A wave is changing from medium A to medium B and it travels slower in medium B than in medium A. If I draw a normal line from the boundary to the wavefront the wavefront in medium B will move away from the normal? I attached to make it more clear

Furthermore, I got this question:

The angle between the wavefronts and the interface in region A is 60. The refractive index anb is 1.4.

What I did was: if 60 between wavefront and boundary, 30 degrees between ray and boundary: sin 30 / sin r = 1.4. I then got between the wavefront and the boundary 69.1 degrees but it is wrong, the mark scheme got 38 degrees. Can anyone help me?

Thanks

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2. Sep 4, 2011

### Staff: Mentor

It looks like they were specifying the angle with respect to the normal to the surface. (They probably mixed them up.)

Edit: On closer inspection, there's nothing wrong with the diagram or problem statement.

Last edited: Sep 4, 2011
3. Sep 4, 2011

### Peter G.

Um, sorry, I am still a bit confused in relation to how they did it. I attached the question since I didn't word it very clearly in my first post:

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4. Sep 4, 2011

### Staff: Mentor

Here's the problem:
What you need in Snell's law is the angle between the ray and the normal.

5. Sep 4, 2011

### Peter G.

Oh, sorry, that was a typo I guess... In the diagram, the angle between the wavefront and the normal is 60 degrees. So I did 90-60 to get the angle between the ray and the normal. I did sin 30 / 1.4 while the mark scheme did sin 60 1.4. So the mark scheme is wrong?

6. Sep 4, 2011

### Staff: Mentor

No, the diagram is quite clear. The angle between the wavefront and the boundary is 60 degrees.

7. Sep 4, 2011

### Peter G.

Ah, never mind, I got it now. Thanks! But if you wouldn't mind, could you help me with the second part of the question I linked? I got the first part but I am confused as to what the marking scheme says:

position of any one minimum closer to centre / minima closer together;
frequency increased so wavelength decreased / correct explanation in terms of double-slit equation;

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8. Sep 4, 2011

### Staff: Mentor

I knew you'd figure it out.
Are you sure you linked it? (See next post--you were still linking when I wrote this.)

9. Sep 4, 2011

### Staff: Mentor

That's the trick. Do you realize that since the speed of the wave is fixed, increasing the frequency must decrease the wavelength? And how will a smaller wavelength affect the pattern? What does the double-slit equation tell you?

10. Sep 4, 2011

### Peter G.

Sorry Doc Al, I updated the post a few seconds after I posted it. I am tired, I shouldn't be studying until now so I am making those frustrating stupid mistakes But this is the last question and then I am done!

11. Sep 4, 2011

### Staff: Mentor

Look above!

12. Sep 4, 2011

### Peter G.

Ah, the distance between the first destructive interference (half a wavelength) to the center will decrease?

13. Sep 4, 2011

### Staff: Mentor

Yes. Smaller wavelength means that a given maximum or minimum will be closer to the center.