Solving Antiphase Wave Homework: Wavelength 633nm

[ravsterphysics]

Homework Statement

Homework Equations

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) )

 

[mfb]

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?

(If antiphase then the path difference is (n+0.5)(lambda) )

That formula is useful.

 

[ravsterphysics]

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.

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?

 

[mfb]

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?

 

[Cutter Ketch]

(If antiphase then the path difference is (n+0.5)(lambda) )

Dude, that's the answer.

 

[Cutter Ketch]

Dude, that's the answer.

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