In a Young's double-slit experiment,the slits | vedclass

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(A) Given:Distance between the slits,$d = 0.28 \; mm = 0.28 	imes 10^{-3} \; m$Distance between the slits and the screen,$D = 1.4 \; m$Distance betwe

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$600$

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(A) Given:Distance between the slits,$d = 0.28 \; mm = 0.28 	imes 10^{-3} \; m$Distance between the slits and the screen,$D = 1.4 \; m$Distance between the central bright fringe and the fourth bright fringe $(n = 4)$,$y_4 = 1.2 \; cm = 1.2 	imes 10^{-2} \; m$The formula for the position of the $n^{th}$ bright fringe is given by $y_n = n \lambda \frac{D}{d}$.Rearranging for wavelength $\lambda$:$\lambda = \frac{y_n \cdot d}{n \cdot D}$Substituting the values:$\lambda = \frac{1.2 	imes 10^{-2} 	imes 0.28 	imes 10^{-3}}{4 	imes 1.4}$$\lambda = \frac{0.336 	imes 10^{-5}}{5.6}$$\lambda = 0.06 	imes 10^{-5} \; m = 6 	imes 10^{-7} \; m$Converting to nanometers $(1 \; m = 10^9 \; nm)$:$\lambda = 6 	imes 10^{-7} 	imes 10^9 \; nm = 600 \; nm$.Therefore,the wavelength of the light used is $600 \; nm$.

In a Young's double-slit experiment,the slits are separated by $0.28 \; mm$ and the screen is placed $1.4 \; m$ away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2 \; cm$. Determine the wavelength of light used in the experiment (in $nm$).

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