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A beam of light consisting of wavelengths 6000 `Å` and 4500 `Å` is used in a YDSE with D = 1 m and d = 1 mm. Find the least distance from the central maxima, where bright fringes due to the two wavelengths coincide.

Text Solution

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`beta_(1)=(lambda_(1)D)/(d)=(6000xx10^(-10)xx1)/(10^(-3))=0.6 mm`
`beta_(2)=(lambda_(2)D)/(d)=0.45 mm`
Let `n_(1)th` maxima of `lambda_(1)` and `n_(2)th` maxima of `lambda_(2)` coincide at a position `y.`
then, `y=n_(1) beta_(1)=n_(2) beta_(2)= LCM` of `beta_(1)` and `beta_(2)`
`rArr y=1.8 mm`
At this point `3rd` maxima for `6000 A & 4th` maxima for `4500 A` coincide
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