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Wave Optics

Question

In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a. Justify this by suitably dividing the slit to bring out the cancellation.

Solution

Let the slit width 'a' be divided into n equal parts of width ' a' ' so that,

a' = \frac{a}{n}

i.e.,

a = na'

Then,
Angle,

θ = \frac{nλ}{a} = \frac{nλ}{na'}

i.e.,

θ = \frac{λ}{a'}

At this derived angle, each slit will make first diffraction minimum. Hence, the resultant intensity for all slits will be zero at an angle of

\frac{nλ}{a}