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A source of light having wavelength 5500a.u is falls normally on slit a width of 22*1/100000cm.calculate the angular seperation of the first two minima on

either side of central maxima.
What's the separation between screen and source?
Oh! Sorry it isn't needed. My mistake.


You need the fringe length ß = λD/d. And angular separation is θ = ß/D = λ/d = (55/22)/(10) = 0.25 radians
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You should write the wavelength as °A  (Angstroms)  and not as a.u.  the unit a.u. is used in astronomy and it is very large.

In the single slit experiment of Young, we have the angle Ф of the minimum intensities on the screen with respect to the perpendicular bisector of the slit as:

     sin Ф ≈ Ф ≈ n λ / a ,  n = +- 1, +- 2, +- 3,.... 
                   where λ = wavelength of the light wave of single frequency
                             a = single slit width

     hence the angular separation of the first two minima = 2λ/a - λ/a = λ/a
                              = 5500 * 10⁻⁸ cm / 22 * 10⁻⁵ cm
                              = 0.25 radians = 14.32 degrees
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