NEET-XII-Physics
17: Light Waves
- #32The line-width of a bright fringe is sometimes defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum. Find the line-width of a bright fringe in a Young’s double slit experiment in terms of
λ, d and D where the symbols have their usual meanings.Ans : Given:
Separation between two slits = d
Wavelength of the light = `` \lambda ``
`` ``
Distance of the screen = `` D``
Let Imax be the maximum intensity and I be half the maximum intensity at a point at a distance y from the central point.
So, `` I={a}^{2}+{a}^{2}+2{a}^{2}\,\mathrm{\,cos\,}\varphi ``
Here, `` \varphi `` is the phase difference in the waves coming from the two slits.
So, `` I=4{a}^{2}{\,\mathrm{\,cos\,}}^{2}\left(\frac{\varphi }{2}\right)``
`` \Rightarrow \frac{I}{{I}_{\,\mathrm{\,max\,}}}=\frac{1}{2}``
`` \Rightarrow \frac{4{a}^{2}{\,\mathrm{\,cos\,}}^{2}\left({\displaystyle \frac{\varphi }{2}}\right)}{4{a}^{2}}=\frac{1}{2}``
`` \Rightarrow {\,\mathrm{\,cos\,}}^{2}\left(\frac{\varphi }{2}\right)=\frac{1}{2}``
`` \Rightarrow \,\mathrm{\,cos\,}\left(\frac{\varphi }{2}\right)=\frac{1}{\sqrt{2}}``
`` \Rightarrow \frac{\,\mathrm{\,\varphi \,}}{2}=\frac{\,\mathrm{\,\pi \,}}{4}``
`` \Rightarrow \varphi =\frac{\,\mathrm{\,\pi \,}}{2}``
`` \,\mathrm{\,Corrosponding\,}\,\mathrm{\,path\,}\,\mathrm{\,difference\,},∆x=\frac{1}{4}``
`` \Rightarrow y=\frac{∆xD}{d}=\frac{\lambda D}{4d}``
The line-width of a bright fringe is defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum.
So, line-width = 2y
`` =2\frac{D\lambda }{4d}=\frac{D\lambda }{2d}``
Thus, the required line width of the bright fringe is `` \frac{D\lambda }{2d}``.
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