NEET-XII-Physics
17: Light Waves
- #19A Young’s double slit apparatus has slits separated by 0⋅28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by red light
λ=700 nm in vacuum. Find the fringe-width of the pattern formed on the screen.Ans : Given:
Separation between two slits, `` d=0.28\,\mathrm{\,mm\,}=0.28\times {10}^{-3}\,\mathrm{\,m\,}``
`` ``
Distance between screen and slit (D) = 48 cm = 0.48 m
Wavelength of the red light, `` {\lambda }_{a}=700\,\mathrm{\,nm\,}\,\mathrm{\,in\,}\,\mathrm{\,vaccum\,}=700\times {10}^{-9}\,\mathrm{\,m\,}``
Let the wavelength of red light in water = `` {\lambda }_{\omega }``
We known that refractive index of water (μw =4/3),
μw = `` \frac{\,\mathrm{\,Speed\,}\,\mathrm{\,of\,}\,\mathrm{\,light\,}\,\mathrm{\,in\,}\,\mathrm{\,vacuum\,}}{\,\mathrm{\,Speed\,}\,\mathrm{\,of\,}\,\mathrm{\,light\,}\,\mathrm{\,in\,}\,\mathrm{\,the\,}\,\mathrm{\,water\,}}``
`` \,\mathrm{\,So\,},{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,w\,}}=\frac{{v}_{a}}{{v}_{\omega }}=\frac{{\lambda }_{a}}{{\lambda }_{\omega }}``
`` \Rightarrow \frac{4}{3}=\frac{{\lambda }_{a}}{{\lambda }_{\omega }}``
`` \Rightarrow {\lambda }_{\omega }=\frac{3{\lambda }_{a}}{4}=\frac{3\times 700}{4}=525\,\mathrm{\,nm\,}``
So, the fringe width of the pattern is given by
`` \,\mathrm{\,\beta \,}=\frac{{\lambda }_{\omega }D}{d}``
`` =\frac{525\times {10}^{-9}\times \left(0.48\right)}{\left(0.28\right)\times {10}^{-3}}``
`` =9\times {10}^{-4}=0.90\,\mathrm{\,mm\,}``
Hence, fringe-width of the pattern formed on the screen is 0.90 mm.
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