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The Separation Between the Consecutive Dark Fringes in a Young'S Double Slit Experiment is 1.0mm. the Screen is Placed at a Distance of 2.5m from the Slits and the Separation Between the Slits Is1.0mm - Physics

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प्रश्न

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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उत्तर

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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पाठ 17: Light Waves - Exercise [पृष्ठ ३८१]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 17 Light Waves
Exercise | Q 6 | पृष्ठ ३८१

संबंधित प्रश्‍न

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(C) equal intensity\

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  1. Find the minimum value of d so that there is a dark fringe at O.
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  3. Find the fringe-width.

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  1. the distance of the second bright fringe from the central maximum for wavelength 500 nm, and
  2. the least distance from the central maximum where the bright fringes due to both wavelengths coincide.

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  1. The angular separation of the fringes.
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