The sin theta = .1 is a typo
Yes, I got 2.87 degrees in the line above.
I don't know where you got sin theta = .1 from??
Look if I use m = 0
and the formula I used then I get L = 500m
when I do arctan(25/500) I get 2.86 degrees.
4 answers
Can you also explain to me why it would be m=0 and not m=1?
Thank you so much for all the help!
Thank you so much for all the help!
I got L = 500 from
y'm = (m+1/2) (lamda L/d)
25 = (0 + 1/2) (5L/50)
L = 500
y'm = (m+1/2) (lamda L/d)
25 = (0 + 1/2) (5L/50)
L = 500
m = 0 because it asked for the first destructive interference. Interference happens when m = 0, +/-1 , +/-2, +/-3 ....etc
m=0 is the first time that the waves arrive half a wavelength apart.
d sin theta = (1/2) lambda
when m = 0
Ok. I see what you did. It is not really a two source interference problem although the effect is the same.
m=0 is the first time that the waves arrive half a wavelength apart.
d sin theta = (1/2) lambda
when m = 0
Ok. I see what you did. It is not really a two source interference problem although the effect is the same.