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Question
In Young's double slit experiment using monochromatic light of wavelength 600 nm, 5th bright fringe is at a distance of 0·48 mm from the centre of the pattern. If the screen is at a distance of 80 cm from the plane of the two slits, calculate:
(i) Distance between the two slits.
(ii) Fringe width, i.e. fringe separation.
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Solution
(i) `lambda = 600nm = 600/1000000 mm`
`lambda =6 xx 10^-4 `mm
D = Distance of slits from the screen = 80cm
⇒ D = 800 mm
`beta` = Fringe - width
`5beta = 0.48` mm
∴ `beta = 0.48/5`
(ii) `beta = 0.096` mm
d = distance between 2 slits
` beta = (lambdaD)/d`
`0.48/5 = (6xx10^-4 xx 8xx10^2)/d`
⇒ `d = (6 xx8xx10^-2 xx5xx10^2)/48 = 5` mm
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