JEE Mains · Physics · STD 12 - 10. Wave optics
A Young's doublc\(-\)slit experiment is performed using monochromatic light of wavelength \(\lambda\). The intensity of light at a point on the screen, where the path difference is \(\lambda,\) is \(K\) units. The intensity of light at a point where the path difference is A \(\frac{\lambda}{6}\) is given by \(\frac{n K}{12},\) where \(n\) is an integer. The value of \(n\) is\(......\)
- A \(9\)
- B \(12\)
- C \(15\)
- D \(5\)
Answer & Solution
Correct Answer
(A) \(9\)
Step-by-step Solution
Detailed explanation
\(I _{\max }= k\) \(I _{1}= I _{2}= K / 4\) \(\Delta x =\lambda / 6 \Rightarrow \Delta \phi=\pi / 3\) \(I = I _{1}+ I _{2}+2 \sqrt{ I _{1} I _{2}} \cos \phi\) \(I =\frac{ K }{4}+\frac{ K }{4}+2 \times \frac{ K }{4} \frac{1}{2}\)…
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