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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 cm