NEET-XII-Physics
17: Light Waves
- #16-aWhat would be the fringe-width? (b) At what distance from the centre will the first maximum be located?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\,}`` (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-bAt 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: