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(A) The position of the $n^{th}$ bright fringe (maxima) is given by the formula:$y_n = \frac{n D \lambda}{d}$Rearranging the formula to solve for wave

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$4000 \ \mathring{A}$

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(A) The position of the $n^{th}$ bright fringe (maxima) is given by the formula:$y_n = \frac{n D \lambda}{d}$Rearranging the formula to solve for wavelength $\lambda$:$\lambda = \frac{y_n d}{n D}$Given values:$n = 3$$y_n = 1.2 \ mm = 1.2 	imes 10^{-3} \ m$$D = 1 \ m$$d = 1 \ mm = 1 	imes 10^{-3} \ m$Substituting these values into the formula:$\lambda = \frac{(1.2 	imes 10^{-3} \ m) 	imes (1 	imes 10^{-3} \ m)}{3 	imes 1 \ m}$$\lambda = \frac{1.2 	imes 10^{-6}}{3} \ m$$\lambda = 0.4 	imes 10^{-6} \ m = 4 	imes 10^{-7} \ m$Converting to $\mathring{A}$s $(\mathring{A})$, where $1 \ \mathring{A} = 10^{-10} \ m$:$\lambda = 4000 	imes 10^{-10} \ m = 4000 \ \mathring{A}$

The distance between two coherent sources is $1 \ mm$. The screen is placed at a distance of $1 \ m$ from the sources. If the distance of the third bright fringe is $1.2 \ mm$ from the central fringe, the wavelength of light used is:

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