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(D) The condition for the $n^{th}$ maximum in a Young's double slit experiment is given by $d \sin 	heta = n \lambda$,where $d$ is the slit separatio

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$\sin^{-1}(3 	imes 10^{-6})$

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(D) The condition for the $n^{th}$ maximum in a Young's double slit experiment is given by $d \sin 	heta = n \lambda$,where $d$ is the slit separation,$	heta$ is the angular position,$n$ is the order of the maximum,and $\lambda$ is the wavelength of light.Given: $d = 0.589 \ m$,$\lambda = 589 \ nm = 589 	imes 10^{-9} \ m$,and $n = 3$.Substituting these values into the formula: $\sin 	heta = \frac{n \lambda}{d} = \frac{3 	imes 589 	imes 10^{-9}}{0.589}$.Since $0.589 = 589 	imes 10^{-3}$,we have $\sin 	heta = \frac{3 	imes 589 	imes 10^{-9}}{589 	imes 10^{-3}} = 3 	imes 10^{-6}$.Therefore,the angular position is $	heta = \sin^{-1}(3 	imes 10^{-6})$.

In the Young's double slit experiment with sodium light,the slits are $0.589 \ m$ apart. The angular separation of the third maximum from the central maximum will be (given $\lambda = 589 \ nm$):

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