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

Vedclass · Accepted Answer

$5280$

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(C) For a single slit diffraction pattern, the condition for the $n^{th}$ secondary maximum is given by $a \sin 	heta = (n + 1/2) \lambda$, where $a$ is the slit width, $	heta$ is the angle of diffraction, and $\lambda$ is the wavelength.Since $	heta$ is very small, $\sin 	heta \approx 	an 	heta = y/D$, where $y$ is the distance from the center and $D$ is the distance to the screen.For the second maximum, $n = 2$, so the condition becomes $a(y/D) = (2 + 1/2) \lambda = 2.5 \lambda$.Given: $a = 0.66 	ext{ mm} = 0.66 	imes 10^{-3} 	ext{ m}$, $D = 1.4 	ext{ m}$, $y = 2.8 	ext{ mm} = 2.8 	imes 10^{-3} 	ext{ m}$.Substituting these values: $(0.66 	imes 10^{-3}) 	imes (2.8 	imes 10^{-3} / 1.4) = 2.5 \lambda$.$(0.66 	imes 10^{-3}) 	imes (2 	imes 10^{-3}) = 2.5 \lambda$.$1.32 	imes 10^{-6} = 2.5 \lambda$.$\lambda = (1.32 	imes 10^{-6}) / 2.5 = 0.528 	imes 10^{-6} 	ext{ m} = 5280 	imes 10^{-10} 	ext{ m} = 5280 	ext{ Å}$.

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