PHY 172 Wave Interference Lab

doc

School

St. Augustine's University *

*We aren’t endorsed by this school

Course

242

Subject

Physics

Date

Dec 6, 2023

Type

doc

Pages

6

Uploaded by MagistrateSeaLionMaster941

Report
Justin Alvarado Wave Interference https://phet.colorado.edu/sims/html/wave - interference/latest/wave - interference_en.html In your lab report, be sure to show calculations. And include screenshots. Part 1. Double Slit Interference In the pictures on the last page, the rays are emitted in all directions from the slits, but let’s concentrate on the rays that are emitted in a direction toward a distant screen ( measured from the normal to the barrier). One of these rays has further to travel to reach the screen, and the path difference is given by d sin . Small angle simplification: If is small (<< 1 rad), then sin  (in radians), bright spots occur on the screen = m ; dark spots occur at = ( m + ) . d d As shown below, the angle (measured from the center of the screen) is related to the distance x measured on the screen by tan( )=x/L, where L is the distance from the screen to the source of light (the aperture). If the angle is small (less than a few degrees), then to an excellent approximation, sin( ) tan( ) (in laser aperture screen L  = x/L tan x and at 2 1
x L d radians) so the locations of the interference bright spots are given by = = m . Procedures: A) Set the slit width to 500 nm and slit separation to 1500 nm. Record your slit separation d in Table 1. B) Press the green button on the light generator and generate an interference pattern on the screen. (Again, you should see something like what you see at the top of this page.) C) Pull the measuring tape tool out of the box in the upper right and use it to measure L (using 3500 nm to 4000 nm) , the distance between the slits and the screen. Then measure x the distance from the center of the central bright spot to the center of one of the 1 st order bright spots. Record these values in Table 1. (Be sure to include units!!!) D) Calculate the wavelength of the light λ using the diffraction formula derived in the Background section. Record this value in Table 1. RED -1792.9nm/3676.3nm = .488 * 1500 nm = 731.5 nm VIOLET- 1056 nm / 3676.3 nm = .287 * 1500 nm = 430.9 nm E) Pause the simulation and use the measuring tape tool to measure the wavelength directly. Record this value in Table 1. RED -(731.5 nm – 696.5 nm) / 731.5 nm *100 = 4.78% VIOLET – (430.9 nm – 410 nm) / 430.9 nm *100 = 4.85% F) Calculate the %-error between your calculated and measured values, and record this value in Table 1. G) Adjust the frequency of light and repeat steps B-F. Color of Light Slit Separation d Distance from Slits to Screen L Distance from Central to 1 st Order Bright Spot x Wavelength λ (calculated) Wavelength λ (measured) %-Error Red 1500 nm 3676.3 nm 1792.9 nm 731.5 nm 696.5 nm 4.78% Violet 1500 nm 3676.3 nm 1056 nm 430.9 nm 410 nm 4.85%
Analysis: 1.What happened to the spacing of the bright spots when you increased the wavelength of the light? By increasing the wavelength of the light, we increased the spacing of the bright spots as well. 2.Explain why your answer to #1 occurred. The spacing of the bright spots increased as well because the fringe width is wavelength dependent, as they are proportional to each other. 3.What happens to the interference pattern if d is increased? What if d is decreased? If (d) is increased, then the spacing of the bright spots will decrease and if (d) is decreased, then the spacing of the bright spots will increase. 4. Explain your reasoning for 3. Insert screenshots here to prove your point. This is because the fringe with of the waves inversely depend on (d). Increased slit separation Decreased slit separation 5. Name one of the Applications. List the source of information.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
One of the applications for double-slit interference is used for learning about the nature of light in that it is not a particle, but a wave. https://www.space.com/double-slit-experiment-light-wave-or-particle# 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 path difference 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/2 so that the total electric field is zero. The angle at which this occurs can be seen from the diagram to be D sin θ=λ [ first minimum] The intensity is a maximum at θ= 0 and decreases to zero at the angle given by equation (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 θ m =mλ Procedures: A) Set the slit width to 1600 nm. 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 nm to 4000 nm) , the distance between the slits and the screen. Then measure x the distance from the center of the central bright spot to the 1 st dark spots. Record these values in Table 1. (Be sure to include units!!!) D) Calculate the wavelength of the light λ using the diffraction formula derived in the Background section. Record this value in Table 1. RED- 1375.4 nm / 3799.1 nm = .362 * 1600 nm = 579.3 nm VIOLET – 859.7 nm / 3799.1 nm = .226 * 1600 nm = 362 nm 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. RED- (579.3 nm – 671.9 nm) / 579.3 nm *100 = 15.99 % VIOLET- (362 nm - 393.6 nm) / 362 nm *100 = 8.73 % G) Adjust the frequency of light and repeat steps B-F. Color of Light Slit width D Distance from Slits to Screen L Distance from Central to 1 st Dark Spot x Wavelength λ (calculated) Wavelength λ (measured) %-Error Red 1600 nm 3799.1 nm 1375.4 nm 579.3 nm 671.9 nm 15.99% Violet 1600 nm 3799.1 nm 859.7 nm 362 nm 393.6 nm 8.73%
Analysis: 1.What happened to the patterns when you increased the wavelength of the light? By increasing the wavelength of the light, the pattern will expand. 2.Explain why your answer to #1 occurred. The pattern will expand because a higher frequency, like the violet light example, corresponds with a shorter wavelength. The red example, which has the highest wavelength, has a lower frequency so the pattern expands more. 3.What happens to the diffraction pattern if d is increased? What if d is decreased? If we increase the slit width, the diffraction pattern will become much bigger and if we decrease the slit width, the diffraction pattern will become much smaller. 4. Explain your reasoning for 3. Insert screenshots here to prove your point. Slit distance at 200nm Slit distance at 1600nm 5. Name one of effects of diffraction seen in every day of life. One of the effects of diffraction in everyday life is a CD. The closely spaced tracks on a CD act as a diffraction grating to form a rainbow color when we look at the CD at a certain angle.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help