In class we have derived expressions for bright and dark fringes in Young’s doub

Question

In class we have derived expressions for bright and dark fringes in Young’s
double-slit experiment by assuming that the distance to the screen d was
much larger than the wavelength ? and the slit separation d. There we took
the two paths to be parallel. Suppose this is no longer true (D is comparable
to d) and the paths are not parallel. Calculate the angle ? (measured from
the point midway between the slits, see figure) of the first-order bright fringe.
Hint: take the square of the path length difference, then square again to get
rid of the square root and obtain tan2?.

Explanation / Answer

The angle ? for the interference maxima can be determined from the below expression ? = sin^-1(m?/d) where m = 1,2,3------- If m =1 for first order bright fringe by condition ?= ?^2 then ? = sin^-1(?^2/d)

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