Young's double slit experiment about light

Published on: **Mar 3, 2016**

Published in:
Science

Source: www.slideshare.net

- 1. Young’s Double-Slit Experiment By: Nancy Manhas
- 2. Importance First performed by Thomas Young, the two slit experiment is key to understanding light and in general the microscopic world The reasoning was to determine whether light was a wave or a particle A source of light illuminates two narrow slits These slits project an image on the further screen
- 3. Light as a Particle If light acted as a particle then where ever it passed through, it would form the image of the slit, with the brightest line in the middle – called the central fringe.
- 4. Light as a Wave If light acted as a wave then it would diffract as it passed through the slits This diffraction through the slits, will give way to the opportunity for points on waves to meet crest to crest or crest to trough. Where crest to crest is constructive, and crest to trough is destructive. Constructive waves are light bands Destructive waves are dark bands
- 5. During Young’s experiment it was noticed that once light passed through the slits it would hit the screen with different densities. Each density, has a wavelength that correlates with a certain color on the spectrum. This diffraction is the light interfering with the wave, and adds more insight to the wave-particle duality. However, one needs to note that not only was Young’s experiment a great contributor to science, it was also a great contributor to the idea that the observer’s eyes do not always reflect what is actually happening.
- 6. Calculations A pair of screens are 14.5m apart. A third order fringe is seen on the screen 3.00cm from the central fringe. If the slits were cut were 0.0960cm apart, determine the wavelength of this light.
- 7. Need to understand that fringes are the bright lines at which constructive interference intervenes. The central fringe is the brightest band, with the highest density of photons. The surrounding bands are numbered by n. Therefore, to determine we need to use a mathematical formula based off the Fraunhofer condition, where the distance between slit one and two are compared to fringe of interest. Simply use the following equation: Lambda = distance from central fringe * distance between slits/length from slits to screen * order of fringe
- 8. Simply plug in the numbers: Lambda = (0.03m)(0.000960m)/3(14.5m) Lambda = 6.62 * 10-7m = 662 *10-9m = 662nm. The wavelength is 662 nm long, which could be found in the range of red.