Question
In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of $\text{n}\lambda/\text{a}.$ Justify this by suitably dividing the slit to bring out the cancellation.
✓
Answer
Let the slit width 'a' be divided into n equal parts of width 'a' so that,
$\text{a}'=\frac{\text{a}}{\text{n}}$
i.e., a = na'
Then,
$\text{Angle},\ \theta=\frac{\text{n}\lambda}{\text{a}}=\frac{\text{n}\lambda}{\text{na}'}$
$\text{i.e.},\ \theta=\frac{\lambda}{\text{a}'}$
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 $\frac{\text{n}\lambda}{\text{a}}.$
$\text{a}'=\frac{\text{a}}{\text{n}}$
i.e., a = na'
Then,
$\text{Angle},\ \theta=\frac{\text{n}\lambda}{\text{a}}=\frac{\text{n}\lambda}{\text{na}'}$
$\text{i.e.},\ \theta=\frac{\lambda}{\text{a}'}$
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 $\frac{\text{n}\lambda}{\text{a}}.$
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