Question

in young’s double slit experiment, the 10 t h maximum of wavelength λ 1 is at distance y 1 from its central maximum and the 5 t h maximum of wavelength λ 2 is at a distance y 2 from its central maximum. the ratio y 1 y 2 will be

Answer 1

Given:
Separation between the slits, d=2 mm=2×10−3 m     
Wavelength of the light, λ=600 nm=6×10−7 m
Distance of the screen from the slits, D = 2⋅0 m
Imax=0.20 W/m2For the point at a position y=0.5 cm=0.5×10−2 m,path difference,∆x=ydD.⇒∆x=0.5×10−2×2×10−32         =5×10−6 m

So, the corresponding phase difference is given by
∆ϕ=2π∆xλ=2π×5×10−66×10−7    =50π3=16π+2π3or ∆ϕ=2π3
So, the amplitude of the resulting wave at point y = 0.5 cm is given by
A=a2+a2+2a2 cos (2π3)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√  =a2+a2−a2‾‾‾‾‾‾‾‾‾‾‾‾√ =a 
Similarly, the amplitude of the resulting wave at the centre is 2a.
Let the intensity of the resulting wave at point y = 0.5 cm be I.
Since IImax=A2(2a)2, we have:I0.2=A24a2=a24a2⇒I=0.24=0.05 W/m2
Thus, the intensity at a point 0.5 cm away from the centre along the width of the fringe