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.


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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