The first minima due to a single slit diffrac | vedclass

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(B) For a single slit diffraction,the condition for the $n^{th}$ minima is given by $a \sin 	heta = n \lambda$,where $a$ is the slit width,$	heta$ i

Vedclass · Accepted Answer

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

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(B) For a single slit diffraction,the condition for the $n^{th}$ minima is given by $a \sin 	heta = n \lambda$,where $a$ is the slit width,$	heta$ is the angle of diffraction,and $\lambda$ is the wavelength of light.Given: $n = 1$,$	heta = 30^o$,and $\lambda = 5000 \, \mathring{A} = 5000 	imes 10^{-10} \, 	ext{m} = 5 	imes 10^{-7} \, 	ext{m}$.Substituting the values into the formula: $a \sin 30^o = 1 	imes (5 	imes 10^{-7} \, 	ext{m})$.Since $\sin 30^o = 0.5$,we have $a 	imes 0.5 = 5 	imes 10^{-7} \, 	ext{m}$.$a = \frac{5 	imes 10^{-7}}{0.5} = 10 	imes 10^{-7} \, 	ext{m} = 10^{-6} \, 	ext{m}$.Converting to centimeters: $a = 10^{-6} 	imes 10^2 \, 	ext{cm} = 10^{-4} \, 	ext{cm} = 1.0 	imes 10^{-4} \, 	ext{cm}$.

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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