In young's double slit experiment,using light of wavelength 400 nm,interference fringes of width 'x' are obtained.The wavelength of light is increased to 600 nm and the separation between the slits is halved.If one wants observed fringe width on the screen to be same in the two cases,find the ratio of distance between the screen and the plane of the interfering sources in the two arrangements
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Hey mate,
● Answer-
D'/D = 1/3
● Explaination-
# Given-
λ1 = 400 nm = 4×10^-7 m
λ2 = 600 nm = 6×10^-7 m
d' = d/2
# Solution-
Fringe width in Young's double slit experiment is given by-
x = λD / d
Case 1-
x = 4×10^-7 × D / d
Case 2-
x' = 6×10^-7 × D' / d'
Equating -
x = x'
4×10^-7 × D / d = 6×10^-7 × D' / d/2
D' = 1/3 D
D'/D = 1/3
Hope this is useful...
● Answer-
D'/D = 1/3
● Explaination-
# Given-
λ1 = 400 nm = 4×10^-7 m
λ2 = 600 nm = 6×10^-7 m
d' = d/2
# Solution-
Fringe width in Young's double slit experiment is given by-
x = λD / d
Case 1-
x = 4×10^-7 × D / d
Case 2-
x' = 6×10^-7 × D' / d'
Equating -
x = x'
4×10^-7 × D / d = 6×10^-7 × D' / d/2
D' = 1/3 D
D'/D = 1/3
Hope this is useful...
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