# 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.

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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