27.5 Single slit diffraction  (Page 2/4)

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 $\text{45.0º}$ 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?

Strategy

From the given information, and assuming the screen is far away from the slit, we can use the equation $D\text{sin}\theta =\text{mλ}$ first to find $D$ , and again to find the angle for the first minimum ${\theta }_1$ .

Solution for (a)

We are given that $\lambda =\text{550 nm}$ , $m=2$ , and ${\theta }_2=\text{45.0º}$ . Solving the equation $D\text{sin}\theta =\text{mλ}$ for $D$ and substituting known values gives

Solution for (b)

Solving the equation $D\text{sin}\theta =\text{mλ}$ for $\text{sin}{\theta }_1$ and substituting the known values gives

Thus the angle ${\theta }_1$ is

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 $\text{20.7º}$ on either side of the original beam, for a width of about $\text{41º}$ . The angle between the first and second minima is only about $\text{24º}(\text{45.0º}-\mathrm{20.7º})$ . Thus the second maximum is only about half as wide as the central maximum.