WBJEE · Physics · Wave Optics
In a Young's double slit experiment, the intensity of light at a point on the screen where the path difference between the interfering waves is \(\lambda\), ( \(\lambda\) being the wavelength of light used) is \(\mathrm{I}\). The intensity at a point where the path difference is \(\frac{\lambda}{4}\) will be (assume two waves have same amplitude)
- A zero
- B I
- C 0.5I
- D 0.25I
Answer & Solution
Correct Answer
(C) 0.5I
Step-by-step Solution
Detailed explanation
\(\mathrm{I}^{\prime}=\mathrm{I} \cos ^2 \phi / 2\) \(\phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}, \quad \phi=\frac{2 \pi}{\lambda} \cdot \frac{\lambda}{4}=\frac{\pi}{2}\)…
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