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  4. In YDSE wavelength of light used is 600nm . At a point intensity is 50 % of maximum value. What is the path difference at that point:-

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Q. In YDSE wavelength of light used is $600nm$ . At a point intensity is $50\%$ of maximum value. What is the path difference at that point:-

NTA AbhyasNTA Abhyas 2022

Solution:

$I=I_{max}\left(cos\right)^{2}\left(\frac{\pi }{\lambda } \cdot \Delta x\right)$
$\Rightarrow \frac{I_{max}}{2}=I_{max}\left(cos\right)^{2}\left(\frac{\pi }{\lambda } \cdot \Delta x\right)$
$\Rightarrow \left(cos \frac{\pi }{\lambda } \cdot \Delta x\right)=\frac{1}{\sqrt{2}}\Rightarrow \Delta x=\frac{\lambda }{4}$