NEET-XII-Physics
17: Light Waves
- #31In a Young’s double slit experiment,
λ=500 nm, d=1·0 mm and D=1·0 m. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.Ans : Given:
Separation between the two slits, `` d=1\,\mathrm{\,mm\,}={10}^{-3}\,\mathrm{\,m\,}``
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Wavelength of the light, `` \lambda =500\,\mathrm{\,nm\,}=5\times {10}^{-7}\,\mathrm{\,m\,}``
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Distance of the screen, `` D=1\,\mathrm{\,m\,}``
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Let Imax be the maximum intensity and I be the intensity at the required 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}``
`` \Rightarrow y=\frac{5\times {10}^{-7}\times 1}{4\times {10}^{-3}}``
`` =1.25\times {10}^{-4}\,\mathrm{\,m\,}``
∴ The required minimum distance from the central maximum is `` 1.25\times {10}^{-4}\,\mathrm{\,m\,}``.
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