A Fraunhofer diffraction pattern due to a sin | vedclass

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(B) Given: Slit width $a = 0.3 	ext{ mm} = 0.3 	imes 10^{-3} 	ext{ m}$.Distance of screen $D = 3 	ext{ m}$.Distance of first minima from central m

Vedclass · Accepted Answer

$5500$

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(B) Given: Slit width $a = 0.3 	ext{ mm} = 0.3 	imes 10^{-3} 	ext{ m}$.Distance of screen $D = 3 	ext{ m}$.Distance of first minima from central maximum $x = 5.5 	ext{ mm} = 5.5 	imes 10^{-3} 	ext{ m}$.For a single slit diffraction, the condition for the first minima is given by $a \sin 	heta = \lambda$.For small angles, $\sin 	heta \approx 	an 	heta = \frac{x}{D}$.Therefore, $a \left( \frac{x}{D} ight) = \lambda$.Substituting the values: $\lambda = \frac{ax}{D} = \frac{(0.3 	imes 10^{-3} 	ext{ m}) 	imes (5.5 	imes 10^{-3} 	ext{ m})}{3 	ext{ m}}$.$\lambda = 0.1 	imes 10^{-3} 	imes 5.5 	imes 10^{-3} 	ext{ m} = 0.55 	imes 10^{-6} 	ext{ m} = 5.5 	imes 10^{-7} 	ext{ m}$.Converting to $	ext{ Å}$: $\lambda = 5500 	imes 10^{-10} 	ext{ m} = 5500 	ext{ Å}$.

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