http://www.ck12.org Chapter 21. Physical Optics
FIGURE 21.8
Check Your Understanding
In order for the light to diffract and create interference, the wavelengthλof the light must be
a. Smaller than the distancedbetween the slits.
b. Equal to the distancedbetween the slits.
c. Larger than the distancedbetween the slits.
Answer: The correct answer is A.
Mathematically, we can see that (for constructive interference),
dsinθ=mλ→sinθ=mλd
The order is given asm≥0. Thus, only form=0 is the relationship between the wavelength of the light and the
distance between the slits irrelevant. Otherwise, the wavelength must be less than the distance between the slits
(λ<d)or the sinθwill not have a solution. We will ignore the case whereθ= 90 ◦since the image would never
appear upon the screen.
Illustrative Example 22.2.1
In an experiment, light incident upon two slits 0.013 mm apart produces a bright third-order fringe at an angle of
8. 7 ◦.
a. What is the wavelength and color of the light used in the experiment?
Solution:
dsinθ=mλ→( 1. 3 × 10 −^5 m)(sin 8. 7 ◦) = 3 λ→λ=
( 1. 3 × 10 −^5 m)(sin 8. 7 ◦)
3= 6. 55 × 10 −^7 m→ 6. 60 × 10 −^7 m= 660 nm
λ= 660 nm(red light)b. If the third fringe appears 30.25 cm(Z)from the center fringe(m= 0 ), what is the distanceLbetween the double
slit and the screen? SeeFigure21.7.
Solution:
We use the angle 8. 7 ◦, the lengthL, and the distanceZ. The tangent of the angle (seeFigure21.7) is