Question
A screen is placed at a distance $50\,cm$ from a single slit, which is illuminated with light of wavelength $690\,nm$. If the distance between first and Third minima is $3.00\,mm,$ then what is the width of the slit
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Answer
The position of $n^{t h}$ minima in the diffraction pattern is $x_{n}=\frac{n D \lambda}{d}$
$\therefore x_{3}-x_{1}=(3-1) \frac{D \lambda}{d}=\frac{2 D \lambda}{d}$
$d=\frac{2 D \lambda}{x_{3}-x_{1}}=\frac{2 \times 0.50 \times 6000 \times 10^{-10}}{3 \times 10^{-3}}=2 \times 10^{-4} \mathrm{m}$
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