AP EAMCET · PHYSICS · Wave Optics
In Young's double slit experiment, the intensity at a point where the path difference is \(\frac{\lambda}{6}\) ( 1 being the wavelength of the light used) is \(I\). If \(I_0\) denotes the maximum intensity. \(\frac{\mathrm{I}}{\mathrm{I}_0}\), is equal to
- A \(\frac{1}{\sqrt{2}}\)
- B \(\frac{\sqrt{3}}{2}\)
- C \(\frac{1}{2}\)
- D \(\frac{3}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{4}\)
Step-by-step Solution
Detailed explanation
In YDSE, \(\Delta x=\frac{\lambda}{6}, \mathrm{I}_1=\mathrm{I}_2\) \(\therefore\) Phase difference, \(\Delta \phi=\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}=\frac{\pi}{3}\)…
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