JEE Mains · Physics · STD 12 - 10. Wave optics
In a Young’s double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is \(1/8^{th}\) of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to
- A \(0.74\)
- B \(0.85\)
- C \(0.94\)
- D \(0.80\)
Answer & Solution
Correct Answer
(B) \(0.85\)
Step-by-step Solution
Detailed explanation
\(\Delta x=\frac{\lambda}{8}\) Phase \(|\Delta P|=\frac{2 \pi}{\lambda} \times \frac{\lambda}{8}=\frac{\pi}{4}\) \(\therefore \) \(\mathrm{I}_{\mathrm{res}}=I+I+2I \cos \left(\frac{\mathrm{R}}{4}\right)\) \(=\,2I\left(1+\frac{1}{\sqrt{2}}\right)=2I \times 1.7\)…
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