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\(....\mu m\)
Pick an option to check your answer.
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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