In a Young\'s interference experiment, the two slits are separated by 0.12 mm an
ID: 2018693 • Letter: I
Question
In a Young's interference experiment, the two slits are separated by 0.12 mm and the incident light includes two wavelengths: ?1 = 540 nm (green) and ?2 = 450 nm (blue). The overlapping interference patterns are observed on a screen 1.48 m from the slits.(a) 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 ?1, and m2 is associated with ?2.)
m2/m1 = 1.2 (CORRECT)
(b) 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 maximum
Explanation / Answer
The condition for interference dsin = m for a small amgle sin = tan = y/L (y/L) d = m Then for two patterns are y1 = m11L/d and y2 = m22L/d When they are coincide y1 = y2 m1 1 L/d = m2 2 L/d m1 1 = m2 2 then m1 /m2 = 2/1 = 450 nm / 540 nm = 5/6 (b) The minimum values for m1 = 5 and m2 = 6 The disatnce from central maximum y = m1 1 L/d = (5)(540 nm) (1.42 m) / 0.15*10^-3 m = 0.0255m or 2.5 cmRelated Questions
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