NEET-XII-Physics
17: Light Waves
- #10A source emitting light of wavelengths 480 nm and 600 nm is used in a double-slit interference experiment. The separation between the slits is 0⋅25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.Ans : Given:
Wavelengths of the source of light,
`` {\lambda }_{1}=480\times {10}^{-9}\,\mathrm{\,m\,}\,\mathrm{\,and\,}{\lambda }_{2}=600\times {10}^{-9}\,\mathrm{\,m\,}``
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Separation between the slits, `` d=0.25\,\mathrm{\,mm\,}=0.25\times {10}^{-3}\,\mathrm{\,m\,}``
Distance between screen and slit, `` D=150\,\mathrm{\,cm\,}=1.5\,\mathrm{\,m\,}``
We know that the position of the first maximum is given by
`` y=\frac{\lambda D}{d}``
So, the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths = y2 - y1
`` {y}_{2}-{y}_{1}=\frac{D\left({y}_{2}-{y}_{1}\right)}{d}``
`` \Rightarrow {y}_{2}-{y}_{1}=\frac{1.5}{0.25\times {10}^{-3}}\left(600\times {10}^{-9}-480\times {10}^{-9}\right)``
`` {y}_{2}-{y}_{1}=72\times {10}^{-5}\,\mathrm{\,m\,}=0.72\,\mathrm{\,mm\,}``
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