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In Young's double slit experiment using monochromatic light of wavelength \( \lambda \), the intensity of
light at a point on the screen where path different is \( \lambda \) is \( \mathrm{K} \) units. The intensity of light at a point
where path difference \( \frac{\lambda}{3} \) is
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light at a point on the screen where path different is \( \lambda \) is \( \mathrm{K} \) units. The intensity of light at a point
where path difference \( \frac{\lambda}{3} \) is
Solution:
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Verified Answer
The correct answer is:
\( \frac{K}{4} \)
\( \Delta x=\frac{\lambda}{2 \Pi} \Delta \phi \)
\( \therefore \frac{\lambda}{3}=\frac{\lambda}{2 \Pi} \cdot \Delta \phi \)
\( \therefore \) Phase difference, \( \phi=\frac{2 \Pi}{3} \)
Intensity, \( I=I_{0} \cos ^{2} \phi \)
\( =K \cdot \cos ^{2}\left(\frac{2 \Pi}{3}\right)=K\left(\frac{-1}{2}\right)^{2} \)
\( I=\frac{K}{4} \)
\( \therefore \frac{\lambda}{3}=\frac{\lambda}{2 \Pi} \cdot \Delta \phi \)
\( \therefore \) Phase difference, \( \phi=\frac{2 \Pi}{3} \)
Intensity, \( I=I_{0} \cos ^{2} \phi \)
\( =K \cdot \cos ^{2}\left(\frac{2 \Pi}{3}\right)=K\left(\frac{-1}{2}\right)^{2} \)
\( I=\frac{K}{4} \)
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