JEE Mains · Physics · STD 12 - 10. Wave optics
Interference fringes are observed on a screen by illuminating two thin slits \(1 \,mm\) apart with a light source \((\lambda=632.8\, nm )\). The distance between the screen and the slits is \(100\, cm\). If a bright fringe is observed on a screen at a distance of \(1.27\, mm\) from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to\(....\mu m\)
- A \(1.27\)
- B \(2\)
- C \(2.87\)
- D \(2.05\)
Answer & Solution
Correct Answer
(A) \(1.27\)
Step-by-step Solution
Detailed explanation
\(y=\frac{n D \lambda}{d}\) \(n=\frac{y d}{D \lambda}=\frac{1.27 \times 10^{-3} \times 10^{-3}}{1 \times 632.8 \times 10^{-9}}=2\) Path difference \(\Delta x=n \lambda\) \(=2 \times 632.8 nm\) \(=1265.6 nm\) \(=1.27 \mu m\)
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