In the double-slit experiment of the figure below, the viewing screen is at dist
ID: 1898593 • Letter: I
Question
In the double-slit experiment of the figure below, the viewing screen is at distance D = 4.00 m, point P lies at distance y = 14.0 cm from the center of the pattern, the slit separation d is 4.50 micro m, and the wavelength lambda is 570 nm. The angle theta serves as a convenient locator for P. For D >> d, we can approximate rays r1 and r2 as being parallel, at angle theta to the central axis. Determine where point P is in the interference pattern by giving the phase difference at point P in radians. What is the ratio of the intensity at point P to the intensity at the center of the pattern (Ip / Icenter)?Explanation / Answer
a)
= (2/) d sin = (2/) d (y/D) = (2*3.1416/570e-9)*(4.5e-6)*(14e-2/4) = 1.73615 1.74 radians
b)
I/Icenter = cos2(/2) = cos2(1.73615/2) = 0.418