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