A beam of light consisting of wavelengths 6000 A° and 4500 A° 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.

β1=λ1Dd=6000×10−10×110−3=0.6mmβ2=λ2Dd=0.45mmLet n1th maxima of λ1 and n2th maxima of λ2 coincide at a position y. then, y=n1β1=n2β2=LCM of β1 and β2 ⇒ y = LCM of 0.6 mm and 0.45 mm y = 1.8 mmAt this point 3 rd maxima for 6000A∘&4 th  maxima for 4500 Ao coincide