JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2019
- Medium
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(0.85\)
Step-by-step Solution
Detailed explanation
\(\Delta x=\frac{\lambda}{8}\)
Phase \(|\Delta P|=\frac{2 \pi}{\lambda} \times \frac{\lambda}{8}=\frac{\pi}{4}\)
\(\therefore \) \(\mathrm{I}_{\mathrm{res}}=I+I+2I \cos \left(\frac{\mathrm{R}}{4}\right)\)
\(=\,2I\left(1+\frac{1}{\sqrt{2}}\right)=2I \times 1.7\)
\(\therefore \quad \frac{I_{res }}{I_{\operatorname{man}}}=\frac{2I\times 1.7}{41}=0.85\)
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