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(B) Given: Slit separation $d = 0.2 	ext{ mm} = 2 	imes 10^{-4} 	ext{ m}$.
Distance to screen $D = 2.0 	ext{ m}$.
Order of fringe $n = 3$.
Dist

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$5000 \mathring{A}$

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(B) Given: Slit separation $d = 0.2 	ext{ mm} = 2 	imes 10^{-4} 	ext{ m}$.
Distance to screen $D = 2.0 	ext{ m}$.
Order of fringe $n = 3$.
Distance of $n^{th}$ bright fringe $y_n = 1.5 	ext{ cm} = 1.5 	imes 10^{-2} 	ext{ m}$.
The formula for the position of the $n^{th}$ bright fringe is $y_n = \frac{n \lambda D}{d}$.
Rearranging for wavelength $\lambda$: $\lambda = \frac{y_n d}{n D}$.
Substituting the values: $\lambda = \frac{1.5 	imes 10^{-2} 	imes 2 	imes 10^{-4}}{3 	imes 2.0}$.
$\lambda = \frac{3 	imes 10^{-6}}{6} = 0.5 	imes 10^{-6} 	ext{ m}$.
Converting to $ \mathring{A}$s: $\lambda = 0.5 	imes 10^{-6} 	imes 10^{10} \mathring{A} = 5000 \mathring{A}$.
Thus,the wavelength of light used is $5000 \mathring{A}$.

In a Young's double slit experiment,the slits are separated by $0.2 \text{ mm}$ and the screen is placed $2.0 \text{ m}$ away. The distance between the central bright fringe and the third bright fringe is measured to be $1.5 \text{ cm}$. Determine the wavelength of light used in the experiment.

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