In a diffraction pattern, light of wavelength | vedclass

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(A) For a single slit diffraction, the condition for the $n^{	ext{th}}$ order maximum is given by $a \sin 	heta = (n + 1/2) \lambda$. For small angl

Vedclass · Accepted Answer

$0.26 	imes 10^{-3} \,m$

Vedclass · Answer

(A) For a single slit diffraction, the condition for the $n^{	ext{th}}$ order maximum is given by $a \sin 	heta = (n + 1/2) \lambda$. For small angles, $\sin 	heta \approx 	an 	heta = x/D$, where $x$ is the distance from the center and $D$ is the distance to the screen. Thus, $a(x/D) = (n + 1/2) \lambda$. For the $2^{	ext{nd}}$ order maximum, $n = 2$, so $a = \frac{(2 + 1/2) \lambda D}{x} = \frac{2.5 \lambda D}{x}$. Given: $\lambda = 580 	imes 10^{-9} \,m$, $D = 2.5 \,m$, $x = 14.5 	imes 10^{-3} \,m$. Substituting the values: $a = \frac{2.5 	imes 580 	imes 10^{-9} 	imes 2.5}{14.5 	imes 10^{-3}} \,m$. $a = \frac{3625 	imes 10^{-9}}{14.5 	imes 10^{-3}} \,m = 250 	imes 10^{-6} \,m = 0.25 	imes 10^{-3} \,m \approx 0.26 	imes 10^{-3} \,m$.

In a diffraction pattern, light of wavelength $580 \,nm$ is incident normally on a slit of width '$a$'. The distance between the slit and the screen is $2.5 \,m$ and the distance of the second-order maximum from the center of the screen is $14.5 \,mm$. The value of '$a$' is:

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