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d sin θ = , for m = 0, 1, 1, 2, 2, . size 12{d`"sin"θ=mλ,`m=0,`1,` - 1,`2,` - 2,` dotslow } {}

For fixed λ size 12{λ} {} and m size 12{m} {} , the smaller d size 12{d} {} is, the larger θ size 12{θ} {} must be, since sin θ = / d size 12{"sin"θ=mλ/d} {} . This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance d size 12{d} {} apart) is small. Small d size 12{d} {} gives large θ size 12{θ} {} , hence a large effect.

The figure consists of two parts arranged side-by-side. The diagram on the left side shows a double slit arrangement along with a graph of the resultant intensity pattern on a distant screen. The graph is oriented vertically, so that the intensity peaks grow out and to the left from the screen. The maximum intensity peak is at the center of the screen, and some less intense peaks appear on both sides of the center. These peaks become progressively dimmer upon moving away from the center, and are symmetric with respect to the central peak. The distance from the central maximum to the first dimmer peak is labeled y sub one, and the distance from the central maximum to the second dimmer peak is labeled y sub two. The illustration on the right side shows thick bright horizontal bars on a dark background. Each horizontal bar is aligned with one of the intensity peaks from the first figure.
The interference pattern for a double slit has an intensity that falls off with angle. The photograph shows multiple bright and dark lines, or fringes, formed by light passing through a double slit.

Finding a wavelength from an interference pattern

Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of 10 . 95º size 12{"10" "." "95"°} {} relative to the incident beam. What is the wavelength of the light?

Strategy

The third bright line is due to third-order constructive interference, which means that m = 3 size 12{m=3} {} . We are given d = 0 . 0100 mm size 12{d=0 "." "0100"`"mm"} {} and θ = 10 . 95º size 12{θ="10" "." "95"°} {} . The wavelength can thus be found using the equation d sin θ = size 12{d`"sin"θ=mλ} {} for constructive interference.

Solution

The equation is d sin θ = size 12{d`"sin"θ=mλ} {} . Solving for the wavelength λ size 12{λ} {} gives

λ = d sin θ m . size 12{λ= { {d`"sin"θ} over {m} } } {}

Substituting known values yields

λ = ( 0 . 0100 mm ) ( sin 10.95º ) 3 = 6 . 33 × 10 4 mm = 633 nm. alignl { stack { size 12{λ= { { \( 0 "." "0100"`"mm" \) \( "sin""10" "." "95" rSup { size 8{ circ } } \) } over {3} } } {} #=6 "." "33" times "10" rSup { size 8{ - 4} } `"mm"="633"`"nm" {} } } {}

Discussion

To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did this for visible wavelengths. This analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with λ size 12{λ} {} , so that spectra (measurements of intensity versus wavelength) can be obtained.

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Calculating highest order possible

Interference patterns do not have an infinite number of lines, since there is a limit to how big m size 12{m} {} can be. What is the highest-order constructive interference possible with the system described in the preceding example?

Strategy and Concept

The equation d sin θ = (for m = 0, 1, 1, 2, 2, ) describes constructive interference. For fixed values of d size 12{d} {} and λ size 12{λ} {} , the larger m size 12{m} {} is, the larger sin θ size 12{"sin"`θ} {} is. However, the maximum value that sin θ size 12{"sin"θ} {} can have is 1, for an angle of 90º size 12{"90"°} {} . (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which m size 12{m} {} corresponds to this maximum diffraction angle.

Solution

Solving the equation d sin θ = size 12{d`"sin"θ=mλ} {}  for  m size 12{m} {} gives

m = d sin θ λ . size 12{m= { {d`"sin"θ} over {λ} } } {}

Taking sin θ = 1 size 12{"sin"θ=1} {} and substituting the values of d size 12{d} {} and λ size 12{m} {} from the preceding example gives

m = ( 0 . 0100 mm ) ( 1 ) 633 nm 15.8. size 12{m= { { \( 0 "." "0100"`"mm" \) \( 1 \) } over {"633"`"nm"} } } {}

Therefore, the largest integer m size 12{m} {} can be is 15, or

m = 15 . size 12{m="15"} {}

Discussion

The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations. However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the center. Consequently, not all 15 fringes may be observable.

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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