NEET-XII-Physics
17: Light Waves
- #20A parallel beam of monochromatic light is used in a Young’s double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Slow that if the incident beam makes an angle
θ=sin-1 λ2dwith the normal to the plane of the slits, there will be a dark fringe at the centre P0 of the pattern.Ans : Let the two slits are S1 and S2 with separation d as shown in figure.
The wave fronts reaching P0 from S1 and S2 will have a path difference of S1X = ∆x.
In ∆S1S2X, `` \,\mathrm{\,sin\,}\theta =\frac{{\,\mathrm{\,S\,}}_{1}\,\mathrm{\,X\,}}{{\,\mathrm{\,S\,}}_{1}{\,\mathrm{\,S\,}}_{2}}=\frac{∆x}{d}``
`` \Rightarrow ∆x=d\,\mathrm{\,sin\,}\theta ``
`` ``
Using `` \,\mathrm{\,\theta \,}={\,\mathrm{\,sin\,}}^{-1}\left(\frac{\lambda }{2d}\right)``, we get,
`` \Rightarrow ∆x=d\times \frac{\lambda }{2d}=\frac{\lambda }{2}``
Hence, there will be dark fringe at point P0 as the path difference is an odd multiple of `` \frac{\lambda }{2}``.
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