In the double-slit experiment of the figure below, the viewing screen is at dist
ID: 1508451 • 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 = 18.0 cm from the center of the pattern, the slit separation d is 4.50 µm, and the wavelength is 645 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.
(b) What is the ratio of the intensity at point P to the intensity at the center of the pattern (IP / Icenter)?
Explanation / Answer
a)
tan = y/D
= 18 cm / 4 m
= 18 cm / 400 cm
= 0.045
for small , tan = sin
so,
sin = 0.045
path difference = d*sin
= 4.5µm * 0.045
= 4500 nm * 0.045
= 202.5 nm
phase difference = path difference* 2*pi/lambda
= 202.5 * 2*pi/645
= 1.973 rad
Answer: 1.973 rad
b)
Iy = Io + Io + 2*sqrt(Io)*sqrt(I0)* cos (phase)
= 2Io + 2Io* cos (1.973 rad)
= 1.22 Io
Icentre = 4 Io
Ratio = 1.22/4 = 0.3
Answer: 0.3