The first minima due to a single slit diffrac | vedclass

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(B) For the first minima in single slit diffraction,the condition is given by $a \sin 	heta = n \lambda$,where $n = 1$.Given: $	heta = 30^o$,$\lambd

Vedclass · Accepted Answer

$1.0 	imes 10^{-4} \, cm$

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(B) For the first minima in single slit diffraction,the condition is given by $a \sin 	heta = n \lambda$,where $n = 1$.Given: $	heta = 30^o$,$\lambda = 5000 \, \mathring{A} = 5000 	imes 10^{-10} \, m = 5 	imes 10^{-7} \, m$.Substituting the values into the formula:$a \sin(30^o) = 1 	imes (5 	imes 10^{-7} \, m)$$a 	imes (0.5) = 5 	imes 10^{-7} \, m$$a = 10 	imes 10^{-7} \, m = 10^{-6} \, m$.Converting the width $a$ into centimeters:$a = 10^{-6} \, m = 10^{-6} 	imes 10^2 \, cm = 10^{-4} \, cm$.Thus,the width of the slit is $1.0 	imes 10^{-4} \, cm$.

The first minima due to a single slit diffraction is at $\theta = 30^o$ for a light of wavelength $5000 \, \mathring{A}$. The width of the slit is

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