3.2 Mathematics of interference  (Page 2/5)

Calculating the highest order possible

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

Strategy

The equation $d\text{sin}\theta =m\lambda$ (for $m=0,\text{±}1,\text{±}2,\text{±}3\text{…}$ ) describes constructive interference from two slits. For fixed values of $d\text{and}\lambda$ , the larger m is, the larger $\text{sin}\theta$ is. However, the maximum value that $\text{sin}\theta$ can have is 1, for an angle of $90\text{°}$ . (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find what value of m corresponds to this maximum diffraction angle.

Solution

Solving the equation $d\text{sin}\theta =m\lambda$ for m gives

Taking $\text{sin}\theta =1$ and substituting the values of $d\text{and}\lambda$ from the preceding example gives

Therefore, the largest integer m can be is 15, or $m=15$ .

Significance

The number of fringes depends on the wavelength and slit separation. The number of fringes is very large for large slit separations. However, recall (see The Propagation of Light and the introduction for this chapter) that wave interference is only prominent when the wave interacts with objects that are not large compared to the wavelength. Therefore, if the slit separation and the sizes of the slits become much greater than the wavelength, the intensity pattern of light on the screen changes, so there are simply two bright lines cast by the slits, as expected, when light behaves like rays. We also note that the fringes get fainter farther away from the center. Consequently, not all 15 fringes may be observable.

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