Question

Please do the calculation step by step, so that I can understand the process. Try not to skip any steps. Thank you

Using the raw data (attach below) from Part 2 and your experimental value of the green laser's wavelength from procedure 2, determine the width of the hair (alternatively, you can use wavelength = 532 nm). You might want to average the distance to the fringes on either side of the maximum since they are not identical. If you like, you can also use a different order of bright fringes to do this calculation, or multiple calculations with different fringe orders to refine your hair width result.

Interfernue / Diffinction Part I Diffraction Ceratory . (60 to front board : 9.453 m distance target distance from ono to "m=1 3.33m "mid to m=-1: 3.18m . distance from 532 an it part 1 = cinty) 9.453m Part 2: Hair width (ure e from part 1 , distance for target to front board distane distone to 4sem ta .. ME-1 5.U cm Cha ma 2 15.6 15.7 cm ni-2 nr 3 26.4 - 3 26

Answer 1

Part 1 : N=600 hrives from is grating constant d = N dre [d- nd <= distance from target to Srist band = 9.453m no to n=1 oran hant ordery diffraction as my iida de 2 = distance from ns * zn 10-3 Goo lines/mm Goo lines/m lo - X 560nm 600 d = ! xe is distance of with band from center :n is taken positive only d x 3.33 587.12 nm 600 9.453x/ for 3.33m Now for x = 3.18m d 10-3 3.18 I mean 9.453x1 =573.56 mm Part 2: Since t = 532 nm is given, we may take that valure only hair acts as an obstacle & diffracts the light. Let hair width = d = 2 9.453m order of chiffraction 4 532 nm se corresponding distance nith band from central band. d and = n sme Z d of 0.1 mm da =0.06 mm de a 1x 532x10-9 x 9.453 0.11 mm 4.5*10-2 dai - 1x532 X10-9 x 9.453 5.0X10-2. = 2x532x10-4x9.453 15.6x10-2 = 2 x 5 32x10-949.453 = 0.06 mm 15.7x10-2 3 *532x10-9x9.453 - 0.057mm 26.4810-2 d43= 3x532x 10-9x9.453 26.0x102 d3 -0.058mm 4.45010- _ 0.074 mm d = day=di +du+de+d-2+dstol-s4.45800** 6

The average is taken for d and that is width of the hair = 0.074mm.

As mentioned in the question, the distnace from first band to the m th band could be averaged and then used to calculate the width. The other method mentioned in question is that the calculations are to be done and averaged at the end, and that is done here. Width was cakculated for all the values of m and then averaged at the end.

More the values to taken for average, better are the results.

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