Example 27.4 from College Physics, 27.5 Single Slit Diffraction

Calculating Single Slit Diffraction

Visible light of wavelength 550 nm falls on a single slit and produces its second diffraction minimum at an angle of $45.0º$ relative to the incident direction of the light. (a) What is the width of the slit? (b) At what angle is the first minimum produced?

The schematic shows a single slit to the left and the resulting intensity pattern on a screen is graphed on the right. The single slit is represented by a gap of size d in a vertical line. A ray of wavelength lambda enters the gap from the left, then five rays leave from the gap center and head to the right. One ray continues on the horizontal centerline of the schematic. Two rays angle upward: the first at an unknown angle theta one above the horizontal and the second at an angle theta two equals forty five degrees above the horizontal. The final two rays angle downward at the same angles, so that they are symmetric about the horizontal with respect to the two rays that angle upward. The intensity on the screen is a maximum where the central ray hits the screen, whereas it is a minimum where the angled rays hit the screen.

Figure 27.24 A graph of the single slit diffraction pattern is analyzed in this example.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
From the given information, and assuming the screen is far away from the slit, we can use the equation $D \sin$ first to find $D$ , and again to find the angle for the first minimum $θ_{1}$ .
Solution for (a)
We are given that $λ = 550 nm$ , $m = 2$ , and $θ_{2} = 45.0º$ . Solving the equation $D sin θ = mλ$ for $D$ and substituting known values gives
$\begin{array}{l} D & = & \frac{mλ}{sin θ_{2}} = \frac{2 (550 nm)}{sin 45.0º} \\ = & \frac{1100 \times 10^{- 9}}{0.707} \\ = & 1.56 \times 10^{- 6} . \end{array}$
27.22
Solution for (b)
Solving the equation $D sin θ = mλ$ for $sin θ_{1}$ and substituting the known values gives
$sin θ_{1} = \frac{mλ}{D} = \frac{1 (550 \times 10^{- 9} m)}{1 . 56 \times 10^{- 6} m} .$
27.23

Thus the angle $θ_{1}$ is

$θ_{1} = {sin}^{- 1} 0.354 = 20.7º.$
27.24
Discussion
We see that the slit is narrow (it is only a few times greater than the wavelength of light). This is consistent with the fact that light must interact with an object comparable in size to its wavelength in order to exhibit significant wave effects such as this single slit diffraction pattern. We also see that the central maximum extends $20.7º$ on either side of the original beam, for a width of about $41º$ . The angle between the first and second minima is only about $24º (45.0º - 20.7º)$ . Thus the second maximum is only about half as wide as the central maximum.

0 of 4 steps open

How did it go?

Strategy

Solution for (a)

Solution for (b)

Discussion

Strategy

Solution for (a)

Solution for (b)

Discussion