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(A) For a single slit diffraction pattern,the condition for the $n^{th}$ order secondary maxima is given by $a \sin 	heta = (n + \frac{1}{2}) \lambda

Vedclass · Accepted Answer

$3.534	imes10^{-3}\, rad$

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(A) For a single slit diffraction pattern,the condition for the $n^{th}$ order secondary maxima is given by $a \sin 	heta = (n + \frac{1}{2}) \lambda$,where $n = 1, 2, 3, \dots$.For the first-order secondary maximum,we set $n = 1$.Thus,$a \sin 	heta = (1 + \frac{1}{2}) \lambda = \frac{3}{2} \lambda$.Since $	heta$ is very small,$\sin 	heta \approx 	heta$.Therefore,$	heta = \frac{3 \lambda}{2 a}$.Given: $\lambda = 5890 	imes 10^{-10}\, m$ and $a = 0.25 	imes 10^{-3}\, m$.Substituting the values: $	heta = \frac{3 	imes 5890 	imes 10^{-10}}{2 	imes 0.25 	imes 10^{-3}}$.$	heta = \frac{17670 	imes 10^{-10}}{0.5 	imes 10^{-3}} = 35340 	imes 10^{-7} = 3.534 	imes 10^{-3}\, rad$.

Determine the angular separation between the central maximum and the first-order maxima of the diffraction pattern due to a single slit of width $0.25\, mm$ when light of wavelength $5890\,\mathring{A}$ is incident on it.

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