Question 291: Waves Optics

Question

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

Accepted Answer

$$\begin{aligned}
S_2 P & =\left(D^2+d^2ight)^{1 / 2} \
& =D\left[1+\frac{d^2}{D^2}ight]^{1 / 2}
\end{aligned}$$
$$
\begin{aligned}
&\begin{aligned}
& \mathrm{S}_2 \mathrm{P}=\mathrm{D}\left[1+\frac{1}{2} \frac{\mathrm{~d}^2}{\mathrm{D}^2}ight]^{1 / 2}=\mathrm{D}+\frac{\mathrm{d}^2}{2 \mathrm{D}} \
& \Rightarrow 	ext { Path difference }=\frac{\mathrm{d}^2}{2 \mathrm{D}}
\end{aligned}\
&	ext { For dark fringe, } \frac{\mathrm{d}^2}{2 \mathrm{D}}=(2 \mathrm{n}-1) \frac{\lambda}{2}\
&\begin{aligned}
	herefore\quad& \frac{d^2}{2 D}=(2 n-1) \frac{d^2}{6 D} \
	herefore\quad& 2 n-1=3 \
	herefore\quad& n=2
\end{aligned}
\end{aligned}$$

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Waves Optics

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