# Waves (antiphase)

1. Dec 14, 2016

### ravsterphysics

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
For the first part I know the wavelength of light is (1.53 x 414nm) = 633nm

But for the second part I'm stumped. Since it's 180 degrees then the waves are in antiphase but I don't understand how to calculate the vertical distance?? (If antiphase then the path difference is (n+0.5)(lambda) )

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2. Dec 14, 2016

### Staff: Mentor

Waves going to A and getting reflected there to get back to the emitter take a longer path than waves getting reflected at B. How much longer?
That formula is useful.

3. Dec 17, 2016

### ravsterphysics

I've taken another look but I'm still confused. Can you help out?

Since it's 180 degrees out of phase then that means it's half a wavelength behind so n would be 0.5, right?

4. Dec 17, 2016

### Staff: Mentor

n is always an integer. the "0.5" are added to n already.

Person X and Y both start at the same place. Person X goes to point B and back to the start. Person Y goes to point A, which is a distance d behind point B, and then goes back to the start. What is the difference in the path lengths of person X and Y?

The difference from above is equal to (n+0.5)(lambda) for some integern n. Which integer n leads to the smallest distance d?

5. Dec 18, 2016

### Cutter Ketch

Dude, that's the answer.

6. Dec 18, 2016

### Cutter Ketch

P.s. Don't forget that the light reflected from A covers the distance between A and B twice.