Envision an opaque screen (Sigma) containing a circular hole of radius R. A poin
ID: 2075891 • Letter: E
Question
Envision an opaque screen (Sigma) containing a circular hole of radius R. A point source S lies on the central axis a distance Rho_0 in front of Sigma and an observation point-P lies a distance r_0 beyond Sigma, also on the central axis. If R = 1.00 mm, Rho_0 = 1.00 m, r_0 = 1.00 m, and Lambda_0 = 500 nm, determine how many Fresnel zones will be visible from P and if it will be brightly illuminated or not. Roughly what would the diffraction pattern look like on a vertical screen containing P? Considering the previous problem, suppose we insert an opaque disk of radius R_D at the center of the hole so that the unobstructed region is now an annulus. If R_D = 0.50 mm, determine the ratio of the irradiance at P now (I) to the irradiance without the screen in place (I_u).Explanation / Answer
10.78)
Squared radius up to nth zone= n.r0.
r0= 1m
= 500 x 10-9m
Radius of aperture = R=1mm=10-3m
Number of zones formed within the aperture can be determined as
R2= n.r0.
n= R2/ r0.= 10-3x 10-3/ (1x 500 x 10-9)=2 zones
The two zones being opposite in phase and nearly equal amplitudes, the effect of the zones will can each other and point P will be dark. (d1~ d2, therefore) A1=d1+d2=d1-d1=0.
10.79)
When a disc of radius 0.5mm covers the aperture, number of zones thus obstructed == RD2/ r0.= (0.5 x 10-3) x (0.5 x 10-3)/ (1x 500 x 10-9)= ½ zone.
Half of the first zone is obstructed. There is finite intensity at P.
A2=|½ d1+d2|= |½ d1- d1|=½ d1
A2/A1=
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