Question : In young's double slit experiment, a light of wavelength 500 nm falls on it. Its split separation is 2 mm and distance between plane of slits and screen is 2m. Then find distance of a point on the screen from central maxima where intensity becomes 50% of central maxima.
Doubt by Eshani
Solution :
λ = 500 nm
= 500×10-9 m
d = 2 mm
= 2×10-3 m
D = 2 m
At central Maxima
Path difference (p) = 0
so Phase difference (Φ) = (2π/λ)(p)
= (2π/λ)(0)
Φ = 0
I=4I₀cos²(Φ/2)
I=4I₀cos²(0/2)
I=4I₀(1)
I=4I₀
I' = 50% of I1
I'= 50% of 4I₀
I' = (1/2)×4I₀
I' = 2I₀
4I₀cos²(Φ'/2) = 2I₀
cos²(Φ'/2)=2I₀/4I₀
cos²(Φ'/2)=1/2
cos(Φ'/2)=1/√2
cos(Φ'/2)=cos(π/4)
Φ'/2 = π/4
Φ'=π/2
Path difference (p')=(λ/2π)(π/2)
p'=λ/4
For constructive Interference
p'=xd/D
λ/4=xd/D|
x = λD/4d
x = [500×10-9×2][4×2×10-3]
x = [10-6][8×10-3]
x = 0.125×10-3 m
x = 1.25×10-4 m
