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NEET-XII-Physics

17: Light Waves

with Solutions - page 5
Qstn# iv-16 Prvs-QstnNext-Qstn
  • #16
    A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0⋅50 mm and the separation between the slits is 0⋅12 cm. The refractive index of mica and polystyrene are 1.58 and 1.55, respectively, for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen at a distance one metre away. (a) What would be the fringe-width? (b) At what distance from the centre will the first maximum be located?
    Ans : Given:
    The thickness of the strips = `` {t}_{1}={t}_{2}=t=0.5\,\mathrm{\,mm\,}=0.5\times {10}^{-3}\,\mathrm{\,m\,}``
    Separation between the two slits,`` d=0.12\,\mathrm{\,cm\,}=12\times {10}^{-4}\,\mathrm{\,m\,}``
    The refractive index of mica, μm = 1.58 and of polystyrene, μp = 1.58
    Wavelength of the light, `` \,\mathrm{\,\lambda \,}=590\,\mathrm{\,nm\,}=590\times {10}^{-9}\,\mathrm{\,m\,},``
    Distance between screen and slit, D = 1 m (a) We know that fringe width is given by
    `` \beta =\frac{\lambda D}{d}``
    `` \Rightarrow \beta =\frac{590\times {10}^{-9}\times 1}{12\times {10}^{-4}}``
    `` =4.9\times {10}^{-4}\,\mathrm{\,m\,}`` (b) When both the mica and polystyrene strips are fitted before the slits, the optical path changes by
    `` ∆x=\left({\,\mathrm{\,\mu \,}}_{\,\mathrm{\,m\,}}-1\right)t-\left({\,\mathrm{\,\mu \,}}_{\,\mathrm{\,p\,}}-1\right)t``
    `` =\left({\,\mathrm{\,\mu \,}}_{\,\mathrm{\,m\,}}-{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,p\,}}\right)t``
    `` =\left(1.58-1.55\right)\times \left(0.5\right)\left({10}^{-3}\right)``
    `` =\left(0.015\right)\times {10}^{-3}\,\mathrm{\,m\,}``
    ∴ Number of fringes shifted, n = `` \frac{∆x}{\lambda }``.
    `` \Rightarrow n=\frac{0.015\times {10}^{-3}}{590\times {10}^{-9}}=25.43``
    ∴ 25 fringes and 0.43th of a fringe.
    ⇒ In which 13 bright fringes and 12 dark fringes and 0.43th of a dark fringe.
    So, position of first maximum on both sides is given by
    On one side,
    `` x=\left(0.43\right)\times 4.91\times {10}^{-4}\left(\because \beta =4.91\times {10}^{-4}\,\mathrm{\,m\,}\right)``
    `` =0.021\,\mathrm{\,cm\,}``
    `` ``
    `` ``
    On the other side,
    `` x\text{'}=\left(1-0.43\right)\times 4.91\times {10}^{-4}``
    `` =0.028\,\mathrm{\,cm\,}``
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  • #16-a
    What would be the fringe-width?
    Ans : We know that fringe width is given by
    `` \beta =\frac{\lambda D}{d}``
    `` \Rightarrow \beta =\frac{590\times {10}^{-9}\times 1}{12\times {10}^{-4}}``
    `` =4.9\times {10}^{-4}\,\mathrm{\,m\,}``
  • #16-b
    At what distance from the centre will the first maximum be located?
    Ans : When both the mica and polystyrene strips are fitted before the slits, the optical path changes by
    `` ∆x=\left({\,\mathrm{\,\mu \,}}_{\,\mathrm{\,m\,}}-1\right)t-\left({\,\mathrm{\,\mu \,}}_{\,\mathrm{\,p\,}}-1\right)t``
    `` =\left({\,\mathrm{\,\mu \,}}_{\,\mathrm{\,m\,}}-{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,p\,}}\right)t``
    `` =\left(1.58-1.55\right)\times \left(0.5\right)\left({10}^{-3}\right)``
    `` =\left(0.015\right)\times {10}^{-3}\,\mathrm{\,m\,}``
    ∴ Number of fringes shifted, n = `` \frac{∆x}{\lambda }``.
    `` \Rightarrow n=\frac{0.015\times {10}^{-3}}{590\times {10}^{-9}}=25.43``
    ∴ 25 fringes and 0.43th of a fringe.
    ⇒ In which 13 bright fringes and 12 dark fringes and 0.43th of a dark fringe.
    So, position of first maximum on both sides is given by
    On one side,
    `` x=\left(0.43\right)\times 4.91\times {10}^{-4}\left(\because \beta =4.91\times {10}^{-4}\,\mathrm{\,m\,}\right)``
    `` =0.021\,\mathrm{\,cm\,}``
    `` ``
    `` ``
    On the other side,
    `` x\text{'}=\left(1-0.43\right)\times 4.91\times {10}^{-4}``
    `` =0.028\,\mathrm{\,cm\,}``
    Page No 381: