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Unformatted text preview: f c − 20 k ) 4 f c − 20 k < B ∑ k
sinc 2 4 f c − 20 k < B ∑ Signal Power
0.1 2000 fc 2000 fc 2000 fc Signal Power
0.1 Signal Power
0.1 41. The diffraction of light can be approximately described through the use of the Fourier
transform. Consider an opaque screen with a small slit being illuminated from the left by
a normally-incident uniform plane light wave (Figure E41).
Solutions 6-33 M. J. Roberts - 8/16/04 Diffracting
Screen x1 Propagation
Slit Wavefronts z Figure E41 One-dimensional diffraction of light through a slit πx
is a good approximation for any x1 in the slit, then the electric field strength of
the light striking the viewing screen can be accurately described by If z >> 2
1 j 2πz π 2π −j
x 0 x1
e λ j λz x 02 ∞
E 0 ( x0 ) = K
∫ E1 ( x1 ) e λz dx1
−∞ where E1 is field strength at the diffracting screen, E 0 is field strength at the viewing screen,
K is a constant of proportionality and λ is the wavelength of the light. The integral is a
Fourier transform with different notation. The field strength at the viewing screen can be
j 2π z
E 0 ( x0 ) = K
F E 1 ( t ) f → x0 . jλ z
λz The intensity, I 0 ( x 0 ) , of the light at the viewing screen is the square of the magnitude of the
I( x 0 ) = E 0 ( x 0 ) .
Plot the intensity of light at the viewing screen if the slit width is 1 mm, the
wavelength of light is 500 nm, the distance, z, is 100 m, the constant of proportionality is
10 −3 and the electric field strength at the diffraction screen is 1 .
The electric field exiting the diffraction slit is
E ( x1 ) = rect 1 0.001
Finding the screen electric field using the formula above, Solutions 6-34 M. J. Roberts - 8/16/04 j 2πz
λ π 2
e x0 e λz sinc
E 0 ( x0 ) = 5 × 10 −2 j 50 The intensity is the square of the magnitude of the electric field, x0 sinc 2 5 × 10 −2 x0 ) =
= 4 × 10 −4 sinc 2 (20 x 0...
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This note was uploaded on 06/19/2013 for the course ENSC 380 taught by Professor Atousa during the Spring '09 term at Simon Fraser.
- Spring '09