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.

A beam of light consisting of wavelengths 6000 `Å` and 4500 `Å` is u

1.	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.
Answer» `beta_(1) = (lambda_(1) D)/(d) = (6000 xx 10^(-10) xx 1)/(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) P_(1) = n_(2) P_(2) = LCM` of `beta_(1)` and `beta_(2)` `implies y = LCM` of `0.6 mm` and `0.45mm` or `y = 1.8 mm` At this point,`3^("rd")` maxima for `6000 Å` and `4^("th")` maxima for `4500 Å` coincide.

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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.

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