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In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of $\mathrm{n} \lambda \sqrt{\mathrm{a}}$. Justify this by suitably dividing the slit to bring out the cancellation.
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It is observed that intensity is zero at $\theta=\frac{n \lambda}{a}$ where $a$ is slit width.
Path difference between extreme
beams from the slit $a \sin \theta=\mathrm{n} \lambda$
for $\mathrm{n}=1$ let us divide the whole slit
in two parts, a part difference $\frac{\lambda}{2}$,
exist between the corresponding
waves from each part. These waves
superimpose destructively on the
screen and cancels each other, thus
producing zero intensity.
Similarly zero intensity for $\mathrm{n}=2,3$,
can also be explained.
Path difference between extreme
beams from the slit $a \sin \theta=\mathrm{n} \lambda$
for $\mathrm{n}=1$ let us divide the whole slit
in two parts, a part difference $\frac{\lambda}{2}$,
exist between the corresponding
waves from each part. These waves
superimpose destructively on the
screen and cancels each other, thus
producing zero intensity.
Similarly zero intensity for $\mathrm{n}=2,3$,
can also be explained.
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