A parallel beam of light of wavelength 6000 , | vedclass

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Question

(A) The condition for the first minima in a single-slit diffraction pattern is given by $a \sin 	heta = n \lambda$,where $n = 1$ for the first minima

Vedclass · Accepted Answer

$2 	imes 10^{-3} \, 	ext{rad}$

Vedclass · Answer

(A) The condition for the first minima in a single-slit diffraction pattern is given by $a \sin 	heta = n \lambda$,where $n = 1$ for the first minima.Given:Wavelength $\lambda = 6000 \, \mathring{A} = 6000 	imes 10^{-10} \, 	ext{m} = 6 	imes 10^{-7} \, 	ext{m}$.Slit width $a = 0.3 \, 	ext{mm} = 0.3 	imes 10^{-3} \, 	ext{m} = 3 	imes 10^{-4} \, 	ext{m}$.For small angles,$\sin 	heta \approx 	heta$.Therefore,$	heta = \frac{\lambda}{a} = \frac{6 	imes 10^{-7}}{3 	imes 10^{-4}} = 2 	imes 10^{-3} \, 	ext{rad}$.

$A$ parallel beam of light of wavelength $6000 \, \mathring{A}$ is diffracted by a single slit of width $0.3 \, \text{mm}$. The angular position of the first minima of the diffracted light is:

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