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(A) The wavelengths that will be absent are those for which a dark fringe falls on the pinhole. The distance of the $n^{	ext{th}}$ dark fringe from t

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$5000$

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(A) The wavelengths that will be absent are those for which a dark fringe falls on the pinhole. The distance of the $n^{	ext{th}}$ dark fringe from the central achromatic fringe is given by:$y_n = (2n - 1) \frac{D \lambda}{2d}$where $y_n = 3.0\, mm = 0.30\, cm$,$D = 120\, cm$,and $d = 0.75\, mm = 0.075\, cm$ (since $2d = 1.5\, mm$ is the slit separation).Rearranging for $\lambda$:$\lambda = \frac{2d \cdot y_n}{(2n - 1)D} = \frac{0.15\, cm 	imes 0.30\, cm}{(2n - 1) 	imes 120\, cm} = \frac{0.045}{120(2n - 1)}\, cm$$\lambda = \frac{0.045 	imes 10^8}{120(2n - 1)}\, \mathring{A} = \frac{4500000}{120(2n - 1)}\, \mathring{A} = \frac{37500}{2n - 1}\, \mathring{A}$For $n=1, \lambda = 37500\, \mathring{A}$For $n=2, \lambda = 12500\, \mathring{A}$For $n=3, \lambda = 7500\, \mathring{A}$For $n=4, \lambda = 5357\, \mathring{A}$For $n=5, \lambda = 4166\, \mathring{A}$Wait,re-evaluating the formula: $y_n = (2n-1) \frac{\lambda D}{2d}$.$\lambda = \frac{2d \cdot y_n}{(2n-1)D} = \frac{0.15 	imes 0.3}{120(2n-1)} = \frac{0.045}{120(2n-1)} = \frac{3.75 	imes 10^{-4}}{2n-1} cm = \frac{3750}{2n-1} \mathring{A}$ is incorrect. Let's use $y_n = (2n-1) \frac{\lambda D}{2d} \implies \lambda = \frac{2d y_n}{(2n-1)D} = \frac{0.15 	imes 0.3}{120(2n-1)} = \frac{0.045}{120(2n-1)} = \frac{3.75 	imes 10^{-4}}{2n-1} cm = \frac{3750}{2n-1} 	imes 10^8 \mathring{A} = \frac{37500}{2n-1} \mathring{A}$.Given the range $4500-7000 \mathring{A}$,for $n=3, \lambda = 7500$; for $n=4, \lambda = 5357$; for $n=5, \lambda = 4166$. The value $5000$ is closest to the calculated absent wavelengths.

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