MHT CET · Physics · Wave Optics
In the Young's double slit experiment, the intensity at a point on the screen, where the path difference is \(\lambda(\lambda=\) wavelength \()\) is \(\beta\). The intensity at a point where the path difference is \(\lambda / 3\), will be \(\left.\cos \frac{\pi}{3}=1 / 2\right]\)
- A \(\beta\)
- B \(\beta / 2\)
- C \(\frac{\beta}{4}\)
- D \(\beta / 8\)
Answer & Solution
Correct Answer
(C) \(\frac{\beta}{4}\)
Step-by-step Solution
Detailed explanation
The intensity is given by
\(\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}\)
Maximum intensity \(\beta=4 \mathrm{I}_0\) when \(\phi=0\) when path difference is \(\frac{\lambda}{3}, \phi=\frac{2 \pi}{3}\)
\(\therefore \quad I=\beta \cos ^2 \frac{2 \pi}{3}=\beta\left(-\frac{1}{2}\right)^2=\frac{\beta}{4}\)
\(\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}\)
Maximum intensity \(\beta=4 \mathrm{I}_0\) when \(\phi=0\) when path difference is \(\frac{\lambda}{3}, \phi=\frac{2 \pi}{3}\)
\(\therefore \quad I=\beta \cos ^2 \frac{2 \pi}{3}=\beta\left(-\frac{1}{2}\right)^2=\frac{\beta}{4}\)
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