27.5 Diffraction
27.5 Diffraction
Diffraction is the bending of waves
around obstacles or the edges of an
opening.
The Amount of Diffraction
Diffraction of Light
Diffraction Minima
Dark fringes for single-slit diffraction
Single-slit Diffraction Pattern
EXAMPLE 6: Single-slit
Diffraction
Light passes through a slit and shines on a flat screen that is
located L = 0.40 m away (see Figure 27.24). The width of the slit
is 4.0×10-6 m. The distance between the middle of the central
bright fringe and the first dark fringe is y. Determine the width 2y
of the central bright fringe when the wavelength of the light in a
vacuum is (a) l = 690 nm (red) and (b)l = 410 nm (violet).
Example 6, Page 834
Light passes through a slit and shines on a flat screen that is located L = 0.40
m away (see Figure 27.24). The width of the slit is 4.0×10-6 m. The distance
between the middle of the central bright fringe and the first dark fringe is y.
Determine the width 2y of the central bright fringe when the wavelength of
the light in a vacuum is (a) l = 690 nm (red) and (b)l = 410 nm (violet).
27.6. Resolving Power
The resolving power of an optical instrument is its ability to
distinguish between two closely spaced objects.
Three photographs of an automobile’s headlights taken at
progressively greater distances from the camera.
Diffraction from Circular Opening
The angle θ, in the picture locates the
first circular dark fringe relative to the
central bright region and is given by:
Rayleigh criterion
It is useful to have a criterion for judging
whether two closely spaced objects will be
resolved by an optical instrument. Figure
27.29a presents the Rayleigh criterion for
resolution, first proposed by Lord Rayleigh
(1842–1919):
Two point objects are just resolved when the
first dark fringe in the diffraction pattern of
one falls directly on the central bright fringe
in the diffraction pattern of the other.
Example 7 The Human Eye
Versus the Eagle’s Eye
A hang glider is flying at an altitude of H = 120 m. Green light (wavelength =
555 nm in vacuum) enters the pilot’s eye through a pupil that has a diameter
of D = 2.5 mm. Determine how far apart two point objects must be on the
ground if the pilot is to have any hope of distinguishing between them. An
eagle’s eye has a pupil with a diameter of D = 6.2 mm. Repeat part (a) for an
eagle flying at the same altitude as the glider.