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प्रश्न
Consider a two-slit interference arrangement (Figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of λ such that the first minima on the screen falls at a distance D from the centre O.
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उत्तर
Young's experiment to show interference of light passing through two slits. A pattern of bright and dark areas appears on the screen (as shown in figure (i)).
The condition for destructive interference is `Δx = S_2P - S_1P +- ((2n - 1)/2)lambda` where n = 1, 2, ...
For nth minima to be formed on the screen path difference (Δx) between the rays coming from S1 and S2 must be `((2n - 1)/2)lambda`
The minima will occur when `Δx = S_2P - S_1P = ((2n - 1)/2)lambda` ......(i)
From the given figure,
`S_1P = sqrt((S_1T_1)^2 + (PT_1)^2) = sqrt(D^2 + (D - x)^2)`
And `S_2P = sqrt((S_2T_2)^2 + (T_2P^2)) = sqrt(D^2 + (D + x)^2)`
`T_2P = T_2O + OP = D + x`
And `T_1P = T_1O - OP = D - x`
Hence, `[D^2 + (D + x)^2]^(-1/2) - [D^2 + (D - x)^2]^(1/2) = lambda/2` ......[For first minima n = 1]
If `x = D`
We can write, `[D^2 + 4D^2]^(-1/2) - [D^2 + 0]^(1/2) = lambda/2`
⇒ `[5D^2]^(1/2) - [D^2 + 0]^(1/2) = lambda/2`
⇒ `sqrt(5)D - D = lambda/2` or `D = lambda/(2sqrt(5) - 1)` = 0.404 λ
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