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Questions
Answer in brief:
The distance between two consecutive bright fringes in a biprism experiment using the light of wavelength 6000 Å is 0.32 mm by how much will the distance change if light of wavelength 4800 Å is used?
The distance between two bright fringes in a biprism experiment using light of wavelength 6000 A.U. 0.32 mm. By how much will the distance change, if the light of wavelength 4800 A.U. is used?
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Solution 1
Data: λ1 = 6000 Å = 6 × 10-7 m, λ2 = 4800 Å = 4.8 × 10-7 m, W1 = 0.32 mm = 3.2 × 10-4 m
Distance between consecutive bright fringes,
W = `(lambda "D")/"d"`
For `lambda_1, "W"_1 = (lambda_1"D")/"d"` and ...(1)
For `lambda_2, "W"_2 = (lambda_2"D")/"d"` and ...(2)
`"W"_2/"W"_1 = (lambda_2 "D"//"d")/(lambda_1 "D"//"d") = lambda_2/lambda_1`
`therefore "W"_2 = (lambda_2/lambda_1)"W"_1 = ((4.8 xx 10^-7)/(6 xx 10^-7)) (3.2 xx 10^-4)`
`= (0.8)(3.2 xx 10^-4)`m
`= 2.56 xx 10^-4` m
`therefore triangle "W" = "W"_1 - "W"_2`
= 3.2 × 10-4 m - 2.56 × 10-4 m
= 0.64 × 10-4 m
= 6.4 × 10-5 m
= 0.064 mm
Solution 2
Given:
Distance between consecutive bright fringes,
yA = 0.32 mm = 0.32 × 10-3 m
λA = 6000 A.U. = 6 × 10-7 m, λB = 4800 A.U. = 4.8 × 10-7 m
Let yB be the distance between consecutive bright fringes when wavelength λB is used
To find: Change in distance between the fringes |yA – yB| Formula: yAλB = yBλA
Calculation:
From formula,
∴ yB = `("y"_"A""y"_"B")/(λ_"A") = (0.32 xx 10^-3 xx 4.8 xx 10^-7)/(6 xx 10^-7)`
= 0.256 × 10-3
∴ Change = |yA – yB|
= = |0.320 × 10-3 - 0.256 × 10-3|
= 0.064 × 10-3 m
= 0.064 mm
The change in distance between the fringes is 0.064 mm.
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