MHT CET · Physics · Wave Optics
In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is \(\lambda\) is ' \(I\) '. The intensity at a point where the path difference is \(\lambda / 6\) is \(\left[\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right]\) \([\lambda=\) wavelength of light \(][\cos \pi=-1]\)
- A I
- B \(\frac{3 \mathrm{I}}{4}\)
- C \(\frac{\mathrm{I}}{2}\)
- D \(\frac{\mathrm{I}}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{3 \mathrm{I}}{4}\)
Step-by-step Solution
Detailed explanation
\(I_{max} = I\) \(I' = I_{max} \cos^2\left(\frac{1}{2} \cdot \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6}\right) = I \cos^2\left(\frac{\pi}{6}\right)\)
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