Chapter 38 Diffraction Patterns and Polarization
38.2 Single-Slit Diffraction
• Each portion of the slit acts as a
source of waves (Huygen’s
principle).
• Light from one portion of the slit
can interfere with light from
another portion.
• The resultant intensity on the
screen depends on the direction
.
• Intensity of the single-slit diffraction pattern (see textbook):
I
=I0
sin(asin
/
)
asin
/
2
Where I0 is the intensity at
= 0 (the central maximum), a
is the width of the slit, and
is the wavelength of light.
• The intensity minimums (destructive interference) occur
when a sin
/
= m (m = ±1, ±2, ±3,…)
Therefore, the condition for destructive interference is
sin
=m
a
(m = ±1, ±2, ±3,…)
• In most cases, use sin
y/L to calculate y, the distance on
the screen.
Example: Light of wavelength 580 nm is incident on a slit of
width 0.3 mm. The observing screen is 2m from the slit
(a) Find the positions of the first dark fringes.
= 580 nm = 580 10-9 m,
a = 0.3 mm = 0.3 10-3 m
L = 2 m
The first dark fringes occurs at m = 1,
sin
=
a
When
is small, sin
y/L
y=
L
a=(580 10
9
m)(2 m)
0.310
3
m=3.87 10
3
m
(b) Find the light intensity 2 mm from the center.
