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(A) For Fraunhofer diffraction at a single slit,the condition for the $n^{th}$ minimum is given by $d \sin 	heta = n \lambda$.Here,$d$ is the slit wi

Vedclass · Accepted Answer

$0.001$

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(A) For Fraunhofer diffraction at a single slit,the condition for the $n^{th}$ minimum is given by $d \sin 	heta = n \lambda$.Here,$d$ is the slit width,$\lambda$ is the wavelength of light,and $	heta$ is the angle of diffraction.For the first minimum,we set $n = 1$.Given values are $\lambda = 550 \,nm = 550 	imes 10^{-9} \,m$ and $d = 0.55 \,mm = 0.55 	imes 10^{-3} \,m$.Substituting these values into the formula: $\sin 	heta = \frac{n \lambda}{d} = \frac{1 	imes 550 	imes 10^{-9}}{0.55 	imes 10^{-3}}$.$\sin 	heta = \frac{550 	imes 10^{-9}}{550 	imes 10^{-6}} = 10^{-3}$.Since $	heta$ is very small,$\sin 	heta \approx 	heta$.Therefore,$	heta = 0.001 \,rad$.

What will be the angle of diffraction for the first minimum due to Fraunhofer diffraction with a source of light of wavelength $550 \,nm$ and a slit of width $0.55 \,mm$ (in $,rad$)?

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