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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2020
  • Hard
  • MCQ

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\(....\mu m\)

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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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