Question:
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 is
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 is
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Path difference at a point \(y\) on the screen is \(\Delta x = \frac{yd}{D}\).
Updated On: Nov 5, 2026
- \(1.27\ \mu\mathrm{m}\)
- \(2.05\ \mu\mathrm{m}\)
- \(2.87\) nm
- 2 nm
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The Correct Option is A
Solution and Explanation
\(\Delta x = \frac{y d}{D}\) with \(y = 1.27 \times 10^{-3}\) m, \(d = 1 \times 10^{-3}\) m, \(D = 1\) m.
\(\Delta x = \frac{1.27 \times 10^{-3} \times 10^{-3}}{1} = 1.27 \times 10^{-6}\) m \(= 1.27\ \mu\)m.
(Check: \(\Delta x / \lambda = 1270/632.8 \approx 2\), consistent with a bright fringe.)
Answer: (1) \(1.27\ \mu\mathrm{m}\)
\(\Delta x = \frac{1.27 \times 10^{-3} \times 10^{-3}}{1} = 1.27 \times 10^{-6}\) m \(= 1.27\ \mu\)m.
(Check: \(\Delta x / \lambda = 1270/632.8 \approx 2\), consistent with a bright fringe.)
Answer: (1) \(1.27\ \mu\mathrm{m}\)
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