75. If the moveable mirror in a Michaelson interferometer ismoved 120 microns, h

Question

75. If the moveable mirror in a Michaelson interferometer ismoved 120 microns, how many fringes would be counted for 600nm oflight?    a.100    b.200    c. 400    d. 720 74. A thin layer of oil (n=1.39) is on top of a puddle ofwater (n=1.33). If normally incident 500nm of light is stronglyreflected, what is the minimum thickness of the oil layer?    a. 90nm    b. 125nm    c. 180nm    d. 250nm 75. If the moveable mirror in a Michaelson interferometer ismoved 120 microns, how many fringes would be counted for 600nm oflight?    a.100    b.200    c. 400    d. 720 74. A thin layer of oil (n=1.39) is on top of a puddle ofwater (n=1.33). If normally incident 500nm of light is stronglyreflected, what is the minimum thickness of the oil layer?    a. 90nm    b. 125nm    c. 180nm    d. 250nm

Explanation / Answer

         = 600nm = 600 * 10-9 m                   Distance associated is = 120 microns             We have :   d = m /2                      Number of finges  m = 2d /                                                                                =2* 120 * 10-6 / 600 * 10-9 m                                                                              = 400             Refractive index of oil is: n=1.39                  = 500 nm = 500 * 10-9 m               We have thickness    t = / 4n                                                                    = 500 * 10-9 m / 4 * 1.39                                                                    = 90 nm   Hope this helps u!

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