20.4 Diffraction Examples
ReadTwo-slit diffraction patterns can be understood mathematically based on the concept of path length and the wavelength. For example, the location of bright antinodes and dark nodes in the two-slit interference pattern can be used to determine the exact wavelength of the light causing them. Read pages 687-689 in your textbook for a full derivation of the mathematical relationships. |
The interference pattern produced by diffraction through a double slit can be analyzed using the following equation:
| Quantity |
Symbol |
SI Unit |
|
wavelength |
λ |
m |
|
slit spacing (separation between slits) |
d |
m |
|
path difference* |
n |
none (number of wavelengths) |
|
angle of diffraction (See page 689 in text.) |
( θ ) |
degrees |
* Constructive interference (antinodes − bright fringes) occurs when the path difference is a whole number of wavelengths ( n = 0, ±1, ±2, ...). Thus, for antinodes or bright fringes: n = 1, 2, 3, 4, ....
Destructive interference (nodes − dark fringes) occurs when the path difference is offset by half a wavelength ( n = ±0.5, ±1.5, ±2.5, ...). Thus, for nodes or dark fringes: n = 0.5, 1.5, 2.5, 3.5, ....
Example Problem 1.
Light with a wavelength of 457 nm is shone through two slits separated by 0.20 mm.
a. What is the angle of diffraction to the second bright fringe ( n = 2)?
Given
Required
the angle of diffraction
Analysis and Solution
Paraphrase
The angle of diffraction is 0.26° from the central bright fringe.
The light producing the second order bright fringe, or n = 2 antinode, is diffracted through a 0.26° angle as it exits the two slits. A simulation can be used to verify this calculation and visualize the interference pattern near the n = 2 bright antinode.
Open the Diffraction Grating simulation and set the slit width, wavelength, and slit count as indicated in the following screen shot.
Remember to press enter after inputting each value. An angle of 0.262o should be displayed. Please note that simulation exaggerates the size of the angles to make them more apparent. You can see that the n = 2 region is located 0.26° from the central antinode.
b. What is the angle of diffraction to the second dark fringe?
Given
Remember that destructive interference, dark fringes, have n values that end in .5. The first dark fringe is 0.5, so the second one is 1.5.
Required
the angle of diffraction to the second dark fringe
Analysis and Solution
Paraphrase
The angle of diffraction to the second dark fringe is 0.20°. If you look at the previous screen shots, you will see that the simulation gives the same result for n = 1.5.
Self-CheckAnswer the following self-check (SC) questions then click the "Check your work" bar to assess your response. |
SC 3.
If some light passes through two slits, separated by 0.33 mm, and the first dark fringe is located 0.035° from the central antinode, what is the wavelength of the light? Show the calculation and check your answer with the simulation.
SC 4.
Blue light of wavelength 465 nm is incident on a double slit where the slits are spaced 0.5 mm apart and are 0.05 mm wide. At what angle of diffraction will the fourth order antinodal line appear? Show the calculation and check your answer with the simulation.
Contact your teacher if your answers vary significantly from the answers provided here.
SC 3.
Required
the wavelength of the light
Analysis and Solution
Paraphrase
The wavelength is 4.0 × 10 -7 m.
SC 4.
Given
Required
the angle of diffraction to the fourth order antinodal line
Analysis and Solution
Paraphrase
The angle of diffraction to the fourth order antinodal line is 0.21°.