WAVE OPTICS

a double seperation of 0.01mm is illuminated by white light . a coloured interference pattern is formed on the screen
100cm away. if a pin hole is placed on this screen at a distance of 2mm from the central fringe , the wavelenghth in the visible spectrum
(4000A to 7000A ) which will be absent in the light through the pin hole is
A) 4000A
B) 5000A
C) 6000A
D) 7000A

{ANS-A} detailed solution shall be appreciated...

1 Answers

1
redion ·

i think i have done something wrong or the values have one zero less, plz. check for destructive interference at point P ( pinhole) , the path diff. = delta (P) = (2n-1) lambda / 2 assuming sin theta = theta = tan theta d* ( 2 mm ) / D = delta (P) = ( 2n - 1) lambda /2 ( 4 mm ) * 10^-5 m / 1 m = ( 2n-1 ) lambda 400 Angstrom / ( 2n-1) = lambda , so the missing wavelength is 400 angstrom because only n=1 is a solution as other wavelengths make it out of range of white light

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