In the far-field diffraction pattern of a sin | vedclass

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(C) For a single slit diffraction pattern,the position of the $n^{th}$ minimum is given by $y_n = \frac{n{\lambda}D}{d}$.For the first minimum $(n=1)$

Vedclass · Accepted Answer

${\lambda _1} = 3.5{\lambda _2}$

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(C) For a single slit diffraction pattern,the position of the $n^{th}$ minimum is given by $y_n = \frac{n{\lambda}D}{d}$.For the first minimum $(n=1)$ with wavelength ${\lambda _1}$,the position is $y_1 = \frac{1 \cdot {\lambda _1}D}{d} = \frac{{\lambda _1}D}{d}$.The position of the $m^{th}$ secondary maximum is given by $y'_m = \frac{(2m+1){\lambda}D}{2d}$.For the third maximum $(m=3)$ with wavelength ${\lambda _2}$,the position is $y'_3 = \frac{(2 	imes 3 + 1){\lambda _2}D}{2d} = \frac{7{\lambda _2}D}{2d} = 3.5\frac{{\lambda _2}D}{d}$.Since the first minimum and third maximum coincide,we equate their positions:$\frac{{\lambda _1}D}{d} = 3.5\frac{{\lambda _2}D}{d}$.Therefore,${\lambda _1} = 3.5{\lambda _2}$.

In the far-field diffraction pattern of a single slit under polychromatic illumination,the first minimum with the wavelength ${\lambda _1}$ is found to be coincident with the third maximum at ${\lambda _2}$. So,

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