In a Young\'s interference experiment, the two slits are separated by 0.18 mm an
ID: 2213870 • Letter: I
Question
In a Young's interference experiment, the two slits are separated by 0.18 mm and the incident light includes two wavelengths: Lambda1 = 540 nm (green) and Lambda2 = 450 nm (blue). The overlapping interference patterns are observed on a screen 1.57 m from the slits. Find a relationship between the orders m1 and m2 that determines where a bright fringe of the green light coincides with a bright fringe of the blue light. (The order m1 is associated with Lambda1, and m2 is associated with Lambda2.) m2/m1 = Your response differs from the correct answer by more than 10%. Double check your calculations. Find the minimum values of m1 and m2 such that the overlapping of the bright fringes will occur and find the position of the overlap on the screen. m1 = m2 = Distance = cm from the central maximumExplanation / Answer
L=1.57 m d=0.18 mm (a) to coincide the bright fringes, m1*lambda1*L/d=m2*lambda2*L/d so m2/m1=540/450=1.2 (b) m2=1.2*m1 to get minimum values of m1 and m2,we test for m1 from 1,2,3 etc and see at what value m2 becomes integer. we see at m1=5,m2=6 so final answer: m1=5 m2=6 distance=0.0236 m=2.36 cm