Doubt by Ridhi
Solution :
d = 2λ
We know, in interference
θ≈nλ/d
or
sinθ=nλ/d
sinθ=nλ/2λ
sinθ=n/2
2sinθ=n
n=2sinθ
The value of sinθ fluctuate between -1 and 1 (both included)
So, possible value of n would be 2(-1) to 2(+1)
or
-2 to +2
i.e. -2, -1, 0, 1 and 2.
We know, in interference
θ≈nλ/d
or
sinθ=nλ/d
sinθ=nλ/2λ
sinθ=n/2
2sinθ=n
n=2sinθ
The value of sinθ fluctuate between -1 and 1 (both included)
So, possible value of n would be 2(-1) to 2(+1)
or
-2 to +2
i.e. -2, -1, 0, 1 and 2.
Hence, the maxium number of possible maxima would be 5.
