Question:
In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is
In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is
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Bright fringes coincide when \(n_1\lambda_1=n_2\lambda_2\); find the smallest integers and use \(y=\frac{n\lambda D}{d}\).
Updated On: Nov 11, 2026
- 15.6 mm
- 1.56 mm
- 7.8 mm
- 9.75 mm
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The Correct Option is C
Solution and Explanation
Bright fringe positions: \(y=\frac{n\lambda D}{d}\). They coincide when \(n_1(650)=n_2(520)\Rightarrow\frac{n_1}{n_2}=\frac{520}{650}=\frac{4}{5}\), so the smallest values are \(n_1=4\), \(n_2=5\).
\(y=\frac{4\times650\times10^{-9}\times1.5}{0.5\times10^{-3}}=4\times650\times10^{-9}\times3000=7.8\times10^{-3}\) m \(=7.8\) mm.
Answer: (3) 7.8 mm
\(y=\frac{4\times650\times10^{-9}\times1.5}{0.5\times10^{-3}}=4\times650\times10^{-9}\times3000=7.8\times10^{-3}\) m \(=7.8\) mm.
Answer: (3) 7.8 mm
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