NEET-XII-Physics

17: Light Waves

with Solutions - page 8
Qstn# iv-38 Prvs-QstnNext-Qstn
  • #38
    A glass surface is coated by an oil film of uniform thickness 1.00 × 10-4 cm. The index of refraction of the oil is 1.25 and that of the glass is 1.50. Find the wavelengths of light in the visible region (400 nm - 750 nm) which are completely transmitted by the oil film under normal incidence.
    Ans : Given:
    Wavelength of light used, `` \lambda =400\times {10}^{-9}\,\mathrm{\,to\,}750\times {10}^{-9}\,\mathrm{\,m\,}``
    Refractive index of oil, μoil, is 1.25 and that of glass, μg, is 1.50.
    The thickness of the oil film, `` d=1\times {10}^{-4}\,\mathrm{\,cm\,}={10}^{-6}\,\mathrm{\,m\,},``
    The condition for the wavelengths which can be completely transmitted through the oil film is given by
    `` \,\mathrm{\,\lambda \,}=\frac{2\,\mathrm{\,\mu \,}d}{\left(n+{\displaystyle \frac{1}{2}}\right)}``
    `` =\frac{2\times {10}^{-6}\times \left(1.25\right)\times 2}{\left(2n+1\right)}``
    `` =\frac{5\times {10}^{-6}}{\left(2n+1\right)}\,\mathrm{\,m\,}``
    `` \Rightarrow \,\mathrm{\,\lambda \,}=\frac{5000}{\left(2n+1\right)}\,\mathrm{\,nm\,}``
    Where n is an integer.
    For wavelength to be in visible region i.e (400 nm to 750 nm)
    When n = 3, we get,
    `` \lambda =\frac{5000}{\left(2\times 3+1\right)}``
    `` =\frac{5000}{7}=714.3\,\mathrm{\,nm\,}``
    When, n = 4, we get,
    `` \lambda =\frac{5000}{\left(2\times 4+1\right)}``
    `` =\frac{5000}{9}=555.6\,\mathrm{\,nm\,}``
    When, n = 5, we get,
    `` \lambda =\frac{5000}{\left(2\times 5+1\right)}``
    `` =\frac{5000}{11}=454.5\,\mathrm{\,nm\,}``
    Thus the wavelengths of light in the visible region (400 nm - 750 nm) which are completely transmitted by the oil film under normal incidence are 714 nm, 556 nm, 455 nm.
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