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In Young's double slit experiment, the slits S1 and S2 are 3 mm apart and the screen is placed 1.0 m away from the slits. It is observed that the fourth bright fringe is at a distance of 5 mm from the - Physics

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Question

In Young's double slit experiment, the slits S1 and S2 are 3 mm apart and the screen is placed 1.0 m away from the slits. It is observed that the fourth bright fringe is at a distance of 5 mm from the second dark fringe. Find the wavelength of light used.

Numerical
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Solution

Fringe positions in YDSE:

`y_n^("bright") = n (lambdaD)/d, n = 0, 1, 2, ...`

`y_m^"dark" = (m + 1/2)(lambdaD)/d, m = 0, 1, 2, ...`

Positions of the given fringes:

ybright ​= position of the 4th bright fringe → n = 4

ydark​ = position of the 2nd dark fringe → m = 2

ybright​ = `4 (lambdaD)/d`

`y_"dark" = (2+1/2) (lambdaD)/d = 5/2 (lambdaD)/d`

ybright​ − ydark ​= 5mm = 5 × 10−3 m

`4 (lambdaD)/d - 5/2 (lambdaD)/d = 5xx10^-3`

`(4-5/2) (lambdaD)/d = 5xx10^-3`

`3/2 (lambdaD)/d = 5xx10^-3`

`3/2xx (lambdaxx1)/(3xx10^-3) = 5xx10^-3`

`(3lambda)/2xx3xx10^10-3`

`lambda/2xx10^-3 = 5xx10^-3`

λ = (2 × 10−3) ⋅ (5 × 10−3) = 10 × 10−6 = 1.0 × 10−5m

λ = 1.0 × 10−5m = 10,000 nm

λ = 10,000 nm

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