Part 2. Single Slit Diffraction
If the viewing screen is far away, the rays heading for any point on the screen are essentially parallel. Consider the waves emanating from the upper half and lower half of the slit.
Destructive interference, a dark fringe, occurs if the pathdifference from any point in the upper half of slit and the corresponding points in the lower half of the slit is an integer multiple of λ/2so that the total electric field is zero. The angle at which this occurs can be seen from the diagram to be
D sinθ=λ [first minimum]
The intensity is a maximum at θ= 0and decreasesto zero at the angle given byequation(3).At a larger angle there will be a bright line, but not nearly as bright as the central spot at θ=0. As the path difference becomes an integer multiple of λ/2, there will again be a minimum of zero intensity when
D sin θm=mλ
Procedures:
A)Setthe slit width to 1600nm.
B)Press the green button on the light generator and generate an interference pattern on the screen.)
C) Pull the measuring tape tool out of the box in the upper right and use it to measure L(using 3500 nmto 4000 nm), the distance between the slits and the screen. Then measure xthe distance from the center of the central bright spot to the 1stdarkspots. Record these values in Table 1. (Be sure to include units!!!)
D) Calculate the wavelength of the light λusing the diffraction formula derived in the Background section. Record this value in Table 1.
E) Pause the simulation and use the measuring tape tool to measure the wavelength directly. Record this value in Table 1.
F) Calculate the %-error between your calculated and measured values, and record this value in Table 1.
Analysis:
1.
2.Explain why your answer to #1occurred.
3.What happens to the diffraction pattern if d is increased? What if d is decreased?
4. Explain your reasoningfor 3.Insert screenshots here to prove your point.
5. Name one of effectsof diffractionseen in every day of life.
