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Question
The separation between the consecutive dark fringes in a Young's double slit experiment is 1.0 mm. The screen is placed at a distance of 2.5m from the slits and the separation between the slits is 1.0 mm. Calculate the wavelength of light used for the experiment.
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Solution
Given:-
Separation between consecutive dark fringes = fringe width (β) = 1 mm = 10−3 m
Distance between screen and slit (D) = 2.5 m
The separation between slits (d) = 1 mm = 10−3 m
Let the wavelength of the light used in experiment be λ.
We know that
\[\beta = \frac{\lambda D}{d}\]
\[{10}^{- 3} m = \frac{2 . 5 \times \lambda}{{10}^{- 3}}\]
\[ \Rightarrow \lambda = \frac{1}{2 . 5} {10}^{- 6} m\]
\[= 4 \times {10}^{- 7} m = 400 \text{ nm}\]
Hence, the wavelength of light used for the experiment is 400 nm.
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