NEET-XII-Physics
17: Light Waves
- #6The 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⋅5 m from the slits and the separation between the slits is 1⋅0 mm. Calculate the wavelength of light used for the experiment.Ans : 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}\,\mathrm{\,m\,}=\frac{2.5\times \lambda }{{10}^{-3}}``
`` \Rightarrow \lambda =\frac{1}{2.5}{10}^{-6}\,\mathrm{\,m\,}``
`` =4\times {10}^{-7}\,\mathrm{\,m\,}=400\,\mathrm{\,nm\,}``
Hence, the wavelength of light used for the experiment is 400 nm.
Page No 381: