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Question
Discuss diffraction at single slit and obtain the condition for nth minimum.
Solution
- Let a parallel beam of light fall normally on a single slit AB of width. The diffracted beam falls on a screen kept at a distance. The center of the slit is C. A straight line through C perpendicular to the plane of slit meets the center of the screen at O.
- All the waves start parallel to each other from different points of the slit and interfere at point P and other points to give the resultant intensities. The point P is in the geometrically shadowed region, up to which the central maximum is spread due to diffraction.
- Condition for P to be first minimum:
- Let us divide the slit AB into two half’s AC and CB. Now the width of AC is (a/2). They are called corresponding points.
- The path difference of light waves from different corresponding points meeting at point P and interfere destructively to make it first minimum.
- The path difference δ between waves from these corresponding points is,
δ = `"a"/2 sin theta`
The Condition for P to be first minimum,
`"a"/2 sin theta = lambda/2`
a sin θ = λ (first minimum)
- Condition for P to be nth order minimum:
Dividing the slit into 2n number of (even number of) equal parts the condition for nth order minimum is
`"a"/"2n"` sin θ = `λ/2`
a sin θ = nλ (nth minimum)
Diffraction at single slit
Corresponding points
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