TSny said:Your general method is correct, but your formula for the minima is incorrect. Check your notes or text.
rude man said:Your minima formula is wrong.
No. 2sin(theta) = 0.5 does not yield theta = 30 deg.elemis said:The formulas listed were in my notes.
Okay, for minima : 2λsinθ = mλ m=0.5,1.5,2.5,3.5...
This would give me the required non-integr half wavelengths for a minima, correct ?
EDIT : Hence m =0.5 would mean the first minima is at 30 degrees.
maxima : 2λsinθ = nλ n = 0,1,2,3,4,5...
For n =1 I still get theta to be 90 degrees... I don't see what I'm doing wrong.
rude man said:No. 2sin(theta) = 0.5 does not yield theta = 30 deg.
No. 2sin(theta) = 1 does not yield theta = 90 deg.