Question:
In Young's double slit experiment, fringes of width \(\beta\) are produced on the screen kept at a distance of 1 m from the slit. When the screen is moved away by \(5 \times 10^{-2}\) m, fringe width changes by \(3 \times 10^{-5}\) m. The separation between the slits is \(1 \times 10^{-3}\) m. The wavelength of light used is .................... nm.
In Young's double slit experiment, fringes of width \(\beta\) are produced on the screen kept at a distance of 1 m from the slit. When the screen is moved away by \(5 \times 10^{-2}\) m, fringe width changes by \(3 \times 10^{-5}\) m. The separation between the slits is \(1 \times 10^{-3}\) m. The wavelength of light used is .................... nm.
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Use $\Delta\beta = \lambda \Delta D / d$ to find wavelength from the change in fringe width when the screen distance changes. No need to know the absolute distance.
Updated On: Apr 20, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Fringe width \(\beta = \lambda D/d\). When \(D\) changes by \(\Delta D\), fringe width changes by \(\Delta\beta = \lambda\Delta D/d\).
Step 2: Detailed Explanation:
\[ \lambda = \frac{\Delta\beta \cdot d}{\Delta D} = \frac{3\times10^{-5} \times 1\times10^{-3}}{5\times10^{-2}} = \frac{3\times10^{-8}}{5\times10^{-2}} = 6\times10^{-7} \text{ m} = 600 \text{ nm} \]
Step 3: Final Answer:
Wavelength of light used \(= 600\) nm.
Fringe width \(\beta = \lambda D/d\). When \(D\) changes by \(\Delta D\), fringe width changes by \(\Delta\beta = \lambda\Delta D/d\).
Step 2: Detailed Explanation:
\[ \lambda = \frac{\Delta\beta \cdot d}{\Delta D} = \frac{3\times10^{-5} \times 1\times10^{-3}}{5\times10^{-2}} = \frac{3\times10^{-8}}{5\times10^{-2}} = 6\times10^{-7} \text{ m} = 600 \text{ nm} \]
Step 3: Final Answer:
Wavelength of light used \(= 600\) nm.
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