7.3 Young’s double slit experiment  (Page 3/9)

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 $\text{10}\text{.}\text{95º}$ 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 $m=3$ . We are given $d=0\text{.}\text{0100}\text{mm}$ and $\theta =\text{10}\text{.}\text{95º}$ . The wavelength can thus be found using the equation $d\text{sin}\theta =\mathrm{m\lambda }$ for constructive interference.

Solution

The equation is $d\text{sin}\theta =\mathrm{m\lambda }$ . Solving for the wavelength $\lambda$ gives

Substituting known values yields

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 $\lambda$ , so that spectra (measurements of intensity versus wavelength) can be obtained.

Calculating 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 and Concept

The equation $d\text{sin}\theta =\mathrm{m\lambda }\text{(for}m=\mathrm{0,}\mathrm{1,}-\mathrm{1,}\mathrm{2,}-\mathrm{2,}\dots )$ describes constructive interference. For fixed values of $d$ 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 $\text{90º}$ . (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which $m$ corresponds to this maximum diffraction angle.

Solution

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

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

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

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.