Light of wavelength 580 text{ nm} is incident | vedclass

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(B) For a single slit diffraction pattern, the condition for the $n^{th}$ order secondary maximum is given by: $a \sin 	heta = (n + 1/2) \lambda$, wh

Vedclass · Accepted Answer

$0.25 	imes 10^{-3} 	ext{ m}$

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(B) For a single slit diffraction pattern, the condition for the $n^{th}$ order secondary maximum is given by: $a \sin 	heta = (n + 1/2) \lambda$, where $n = 1, 2, 3, ...$For small angles, $\sin 	heta \approx 	an 	heta = y/D$.Thus, $a(y/D) = (n + 1/2) \lambda$.Given: $\lambda = 580 	imes 10^{-9} 	ext{ m}$, $D = 2.5 	ext{ m}$, $y = 14.5 	imes 10^{-3} 	ext{ m}$, and $n = 2$.Substituting the values: $a 	imes (14.5 	imes 10^{-3} / 2.5) = (2 + 0.5) 	imes 580 	imes 10^{-9}$.$a 	imes (5.8 	imes 10^{-3}) = 2.5 	imes 580 	imes 10^{-9}$.$a = (2.5 	imes 580 	imes 10^{-9}) / (5.8 	imes 10^{-3})$.$a = (1450 	imes 10^{-9}) / (5.8 	imes 10^{-3}) = 250 	imes 10^{-6} 	ext{ m} = 0.25 	imes 10^{-3} 	ext{ m}$.

Light of wavelength $580 \text{ nm}$ is incident normally on a slit of width '$a$'. The distance between the slit and the screen is $2.5 \text{ m}$ and the distance of the second order maximum from the centre of the screen is $14.5 \text{ mm}$ in a diffraction pattern. The value of '$a$' is

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