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(A) For single slit diffraction,the condition for the $n^{th}$ dark fringe is given by $d \sin 	heta = n \lambda$,where $d$ is the slit width,$	heta

Vedclass · Accepted Answer

$6000$

Vedclass · Answer

(A) For single slit diffraction,the condition for the $n^{th}$ dark fringe is given by $d \sin 	heta = n \lambda$,where $d$ is the slit width,$	heta$ is the angular position,and $n$ is the order of the fringe.Given: Slit width $d = 24 	imes 10^{-5} \ cm = 24 	imes 10^{-7} \ m$.Order of fringe $n = 2$.Angular position $	heta = 30^{\circ}$.Substituting the values into the formula: $24 	imes 10^{-7} 	imes \sin(30^{\circ}) = 2 	imes \lambda$.Since $\sin(30^{\circ}) = 0.5$,we have $24 	imes 10^{-7} 	imes 0.5 = 2 \lambda$.$12 	imes 10^{-7} = 2 \lambda$.$\lambda = 6 	imes 10^{-7} \ m$.Converting to $\mathring{A}$: $\lambda = 6 	imes 10^{-7} 	imes 10^{10} \ \mathring{A} = 6000 \ \mathring{A}$.

$A$ light wave is incident normally on a slit of width $24 \times 10^{-5} \ cm$. The angular position of the second dark fringe from the central maxima is $30^{\circ}$. What is the wavelength of light in $\mathring{A}$?

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