In the double-slit experiment of the figure below, the viewing screen is at dist
ID: 2254869 • 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
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 µm, and the wavelength ? is 625 nm. The angle ? serves as a convenient locator for P. For D >> d, we can approximate rays r1 and r2 as being parallel, at angle ? 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 / I center)?Explanation / Answer
The wave length of light ? = 625nm
the seperation between slits d = 4.50?m
the distance between screen D = 4.0m
the postion of point P is y = 14cm
(a) The position of P in terms of phase difference
? = 2?/? dsin?
=[ 2?/(625*10^-9)](4.5*10^-6)(y/D)
since for small angles sin? ? tan? = y/D
= 0.0452*10^3 (14*10^-2/4)
= 1.582 rad
(b) The ratio of intensity
I/I0 = 4cos^2(?/2)
= 4cos^2(1.582/2)
=3.99=4