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In a YDSE arrangement, the distance of s...

In a YDSE arrangement, the distance of screen from the slits is half the distance between the slits. If `'lambda'` is the wavelength of then the value of D such that first minima on the screen falls at a distance D from then centre 'O' is

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As is clear from Fig.
`T_(2)P = T_(2)P + OP = S_(2)C + OP = D + x`
`T_(1)P = OT_(1) - OP = CS_(1) - OP = D - x`
Now, `S_(1)P = sqrt(S_(1)T_(1)^(2) + T_(1)P^(2)) = [D^(2) + (D - x)^(2)]^(1//2)`
and `S_(2)P = sqrt(S_(1)T_(2)^(2) + T_(2)P^(2)) = [D^(2) + (D - x)^(2)]^(1//2)`
Path diff. between the waves reaching `P` from `S_(1)` and `S_(2)`
`= S_(2)P - S_(1)P`
`= [D^(2) + (D + x)^(2)]^(1//2) - [D^(2) + (D - x)^(2)]^(1//2)` ltbegt For first minimum to fall at `P, n = 1`
Path diff. `= S_(2)P - S_(1)P = (1 lambda)/(2)`
or `[D^(2) + (D + x)^(2)]^(1//2) - (D^(2) + (D - x)^(2)]^(1//2) = (lambda)/(2)`.
If `x = D`
`[D^(2) + 4 D^(2)]^(1//2) - D = (lambda)/(2)`
`D[sqrt(5) - 1) = (lambda)/(2) or D(2.236 -1) = lambda//2` or `D = (lambda)/(2.472)`
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