Q10.21 In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of \frac{n\lambda}{a}. Justify this by suitably dividing the slit to bring out the cancellation.

Answers (1)

Let the width of the slit b be divided into n equal parts so that 

b'=\frac{b}{n}

b=b'n

Now,

\theta=\frac{n\lambda}{b}=\frac{n\lambda}{b'n}=\frac{\lambda}{b'}

At this angle, each slit will make the first diffraction minimum. therefore the resultant intensity for all the slits will be zero at the angle of \frac{n\lambda}{b}.

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