In the double-slit experiment of the figure below, the viewing screen is at dist
ID: 1697739 • 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 = 13.0 cm from the center of the pattern, the slit separation d is 4.50 µm, and the wavelength is 650 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.
(a) Determine where point P is in the interference pattern by giving the phase difference at point P in radians.
1
(b) What is the ratio of the intensity at point P to the intensity at the center of the pattern (IP / Icenter)?
2
Explanation / Answer
The wave length of light = 650nm
the seperation between slits d = 4.50m
the distance between screen D = 4.0m
the postion of point P is y = 13cm
(a) The position of P in terms of phase difference
= 2/ dsin
=[ 2/(650*10^-9)](4.5*10^-6)(y/D)
since for small angles sin tan = y/D
= 0.0435*10^3 (13*10^-2/4)
= 0.45 rad = 1.413 rad
(b) The ratio of intensity
I/I0 = 4cos^2(/2)
= 4cos^2(1.413/2)
=2.3