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NEET-XII-Physics

20: dispersion and Spectra

with Solutions - page 2
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  • #2
    A certain material has refractive indices 1.56, 1.60 and 1.68 rfor red, yellow and violet lightespectively. (a) Calculate the dispersive power. (b) Find the angular dispersion produced by a thin prism of angle 6° made of this material. (a) Calculate the dispersive power. (b) Find the angular dispersion produced by a thin prism of angle 6° made of this material.
    Ans : Given:
    Refractive index for red light, μr = 1.56
    Refractive index for yellow light, μy = 1.60
    Refractive index for violet light, μv = 1.68
    Angle of prism, A = 6° (a) Dispersive power `` \left(\omega \right)`` is given by
    `` \omega =\frac{{\mu }_{v}-{u}_{r}}{{\mu }_{y}-1}``
    `` \,\mathrm{\,On\,}\,\mathrm{\,substituting\,}\,\mathrm{\,the\,}\,\mathrm{\,values\,}\,\mathrm{\,in\,}\,\mathrm{\,the\,}\,\mathrm{\,above\,}\,\mathrm{\,formula\,},\,\mathrm{\,we\,}\,\mathrm{\,get\,}:``
    `` \omega =\frac{\left(1.68-1.56\right)}{\left(1.60-1\right)}``
    `` =\frac{0.12}{0.60}=0.2`` (b) Angular dispersion = (μv - μr)A
    =(0.12) × 6° = 0.72°
    Thus, the angular dispersion produced by the thin prism is 0.72°.
    Page No 442: (a) Dispersive power `` \left(\omega \right)`` is given by
    `` \omega =\frac{{\mu }_{v}-{u}_{r}}{{\mu }_{y}-1}``
    `` \,\mathrm{\,On\,}\,\mathrm{\,substituting\,}\,\mathrm{\,the\,}\,\mathrm{\,values\,}\,\mathrm{\,in\,}\,\mathrm{\,the\,}\,\mathrm{\,above\,}\,\mathrm{\,formula\,},\,\mathrm{\,we\,}\,\mathrm{\,get\,}:``
    `` \omega =\frac{\left(1.68-1.56\right)}{\left(1.60-1\right)}``
    `` =\frac{0.12}{0.60}=0.2`` (b) Angular dispersion = (μv - μr)A
    =(0.12) × 6° = 0.72°
    Thus, the angular dispersion produced by the thin prism is 0.72°.
    Page No 442:
  • #2-a
    Calculate the dispersive power.
    Ans : Dispersive power `` \left(\omega \right)`` is given by
    `` \omega =\frac{{\mu }_{v}-{u}_{r}}{{\mu }_{y}-1}``
    `` \,\mathrm{\,On\,}\,\mathrm{\,substituting\,}\,\mathrm{\,the\,}\,\mathrm{\,values\,}\,\mathrm{\,in\,}\,\mathrm{\,the\,}\,\mathrm{\,above\,}\,\mathrm{\,formula\,},\,\mathrm{\,we\,}\,\mathrm{\,get\,}:``
    `` \omega =\frac{\left(1.68-1.56\right)}{\left(1.60-1\right)}``
    `` =\frac{0.12}{0.60}=0.2``
  • #2-b
    Find the angular dispersion produced by a thin prism of angle 6° made of this material.
    Ans : Angular dispersion = (μv - μr)A
    =(0.12) × 6° = 0.72°
    Thus, the angular dispersion produced by the thin prism is 0.72°.
    Page No 442: