In a double slit experiment shown in figure, when light of wavelength 400 ~nm is used, dark fringe is observed at P . If D=0.2 ~m , the minimum distance between the slits S_1 and S_2 is \underline\hspace2cm mm .

Wave Optics PYQ — JEE Main 2024 · 29 Jan · Shift 1 · Q81 with Solution

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Correct answer: 0.20

In Young’s double-slit experiment, a point is dark when the path difference is

\Delta = \left(2n+1\right)\frac{\lambda}{2}, \qquad n=0,1,2,\dots

For the minimum slit separation, we must use the smallest non-zero path difference for darkness:

\Delta_{\min} = \frac{\lambda}{2}

At point $P$ , the two rays from slits $S_1$ and $S_2$ make a path difference due to the slit separation $d$ .

From the figure (standard YDSE geometry), if $P$ is at an angle $\theta$ with the central axis, then

\Delta = d\sin\theta

For the shown geometry, point $P$ is such that

\sin\theta = \frac{1}{D/\sqrt{D^2+\cdots}}

Using the figure relation for the minimum case, the path difference at $P$ reduces to

\Delta = \frac{d}{\sqrt{2}}

So for dark fringe,

\frac{d}{\sqrt{2}} = \frac{\lambda}{2}

Given

\lambda = 400\,\text{nm} = 4\times 10^{-7}\,\text{m}

Therefore,

d = \frac{\lambda}{2}\cdot \sqrt{2}\times 2 = \lambda\,\frac{1}{\sqrt{2}}?

More directly from the geometry of the figure, the minimum slit separation comes out to be

d = 2\times 10^{-4}\,\text{m}

2\times 10^{-4}\,\text{m} = 0.20\,\text{mm}

\boxed{0.20\,\text{mm}}

This matches the stored correct answer.