In Young's double slit experiment, the distance between the two coherent sources is ' d ' and the distance between the source and screen is ' D '. When the wavelength ( $\lambda$ ) of light source used is $\frac{d^2}{3 D}$, then $n^{ ext {th }}$ dark fringe is observed on the screen, exactly in…

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In Young's double slit experiment, the distance between the two coherent sources is ' d ' and the distance between the source and screen is ' D '. When the wavelength ( $\lambda$ ) of light source used is $\frac{d^2}{3 D}$, then $n^{	ext {th }}$ dark fringe is observed on the screen, exactly in front of one of the slits. The value of ' $n$ ' is

Aicharya · Accepted Answer

$\begin{aligned} \mathrm{S}_2 \mathrm{P} & =\left(\mathrm{D}^2+\mathrm{d}^2\right)^{1 / 2} \\ & =\mathrm{D}\left[1+\frac{\mathrm{d}^2}{\mathrm{D}^2}\right]^{1 / 2}\end{aligned}$

Using binomial equation, $\begin{aligned} & \mathrm{S}_2 \mathrm{P}=\mathrm{D}\left[1+\frac{1}{2} \frac{\mathrm{~d}^2}{\mathrm{D}^2}\right]^{1 / 2}=\mathrm{D}+\frac{\mathrm{d}^2}{2 \mathrm{D}} \\ & \Rightarrow \text { Path difference }=\frac{\mathrm{d}^2}{2 \mathrm{D}} \end{aligned}$ For dark fringe, $\frac{\mathrm{d}^2}{2 \mathrm{D}}=(2 \mathrm{n}-1) \frac{\lambda}{2}$ $\begin{aligned} & \therefore \quad \frac{\mathrm{d}^2}{2 \mathrm{D}}=(2 \mathrm{n}-1) \frac{\mathrm{d}^2}{6 \mathrm{D}} \\ & \therefore \quad 2 \mathrm{n}-1=3 \\ & \therefore \quad n=2\end{aligned}$ ~

In Young's double slit experiment, the distance between the two coherent sources is ' d ' and the distance…

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