

Subject:
optical physics constructive interference
Category: Science > Physics Asked by: gugaga List Price: $25.00 
Posted:
04 Aug 2002 14:15 PDT
Expires: 03 Sep 2002 14:15 PDT Question ID: 50551 

Subject:
Re: optical physics constructive interference
Answered By: alienintelligencega on 05 Aug 2002 00:50 PDT Rated: 
Hi guga I'm glad I got a chance to answer this question. It did a good job of knocking the rust off the physics I took. !!!!!!!!!!!!!!!!!! Problem #1a: Equation for the thickness of the paper(t) in terms of the wavelength of the light through the gap between the plates. The thickness of the paper can be determined by the order (number) of fringes from where the plates touch to the edge of the paper, with the wavelength known. The distance the light must travel in the gap is roundtrip. The minimum path length difference is half a wavelength. Thus, t = ((m + 1/2)(wavelength)) / 2 Where t = thickness of paper (unknown) m = number of bright fringes to the paper or the order of bright fringes (unknown) wavelength = 552nm Determine the order (m) of bright fringes The order of fringes (m) is relative to the distance from the point of contact of the glass plates to the edge of the paper by: m = (2t / wavelength)  1/2 Where t = thickness of paper (unknown) m = number of bright fringes to the paper or the order of bright fringes (unknown) wavelength = 552nm !!!!!!!!!!!!!!!!!! Problem #1b: Assume the thickness of the paper in problem #1a (above) is 4x(10^5)m meters, how many bright fringes are observed. The number of bright fringes (m) = twice the thickness of the paper divided by the wavelength minus one half. m = (2t / wavelength)  1/2 where t = 4x(10^5)m and wavelength = 552x(10^9)m So, m = (2 x 4x(10^5)m / 552x(10^9)m)  0.5 = 144.43 rounded we get 144. The first bright fringe occurs at m = 0 so total bright fringes would equal 145 !!!!!!!!!!!!!!!!!! Problem #1c: Explain why there is a dark fringe where the plates touch. There is a dark fringe where the plates touch because of destructive interference at that point. This is caused by the phase shift upon reflection of the wave reflected from the lower plate at a distance of 0. m = (2 * 0 /(552x(10^9)m)  1/2 m = 1/2 halfwavelength phase shift Search Techniques wedge gap two pieces of glass wavelength fringe angle [ ://www.google.com/search?num=20&hl=en&lr=&ie=ISO88591&safe=off&q=wedge+gap+two+pieces+of+glass+wavelength+fringe+angle ] two glass plates separated gap angle wavelength light fringe bright [ ://www.google.com/search?num=20&hl=en&lr=&ie=ISO88591&safe=off&q=two+glass+plates+separated+gap+angle+wavelength+light+fringe+bright ] glass plates gap angle wavelength light fringe [ ://www.google.com/search?num=20&hl=en&lr=&ie=ISO88591&safe=off&q=glass+plates+gap+angle+wavelength+light+fringe&btnG=Google+Search ] [ http://131.128.120.24/PHY205/PHY205HW/exam4/index.html ] [ http://216.239.37.100/search?q=cache:tey7hn7mLL8C:niuhep.physics.niu.edu/~willis/phys251/chapter_27_day_1.html+glass+plates+angle+wavelength+light++fringe&hl=en&ie=UTF8 ] I sure hope this is correct, it looks like I only get one shot at this. I had a nice lil diagram all made up in Adobe Illustrator... but I have no where to upload it to, for you to publicly access it. thanks for asking us this question AI 
gugaga
rated this answer:
I rated you because you were fast in getting the ans. and gave some infor where to look. you are great. also to let you know that I am getting close to BIG 50 so I just don't have that energy to go back to school. thanks 

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