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NEET-XII-Physics

17: Light Waves

with Solutions - page 6
Qstn# iv-25-a Prvs-QstnNext-Qstn
  • #25-a
    Which wavelength(s) will be absent in the light coming from the hole? (b) Which wavelength(s) will have a strong intensity?
    Figure
    Ans : The wavelength(s) will be absent in the light coming from the hole, which will form a dark fringe at the position of hole.
    `` {y}_{n}=\frac{\left(2n+1\right){\lambda }_{n}}{2}\frac{D}{d}`` , where n = 0, 1, 2, ...
    `` \Rightarrow {\lambda }_{n}=\frac{2}{\left(2n+1\right)}\frac{{y}_{n}d}{D}``
    `` =\frac{2}{\left(2n+1\right)}\times \frac{{10}^{-3}\times 0.05\times {10}^{-3}}{0.5}``
    `` =\frac{2}{\left(2n+1\right)}\times {10}^{-6}\,\mathrm{\,m\,}``
    `` =\frac{2}{\left(2n+1\right)}\times {10}^{3}\,\mathrm{\,nm\,}``
    `` \,\mathrm{\,For\,}n=1,``
    `` {\lambda }_{1}=\left(\frac{2}{3}\right)\times 1000=667\,\mathrm{\,nm\,}``
    `` \,\mathrm{\,For\,}n=2,``
    `` {\lambda }_{2}=\left(\frac{2}{5}\right)\times 1000=400\,\mathrm{\,nm\,}``
    Thus, the light waves of wavelength 400 nm and 667 nm will be absent from the light coming from the hole. (b) The wavelength(s) will have a strong intensity, which will form a bright fringe at the position of the hole.
    `` \,\mathrm{\,So\,},{y}_{n}=n{\lambda }_{n}\frac{D}{d}``
    `` \Rightarrow {\lambda }_{n}={y}_{n}\frac{d}{nD}``
    `` \,\mathrm{\,For\,}n=1,``
    `` {\lambda }_{1}={y}_{n}\frac{d}{D}``
    `` ={10}^{-3}\times \left(0.5\right)\times \frac{{10}^{-3}}{0.5}``
    `` ={10}^{-6}\,\mathrm{\,m\,}=1000\,\mathrm{\,nm\,}.``
    But 1000 nm does not fall in the range 400 nm - 700 nm.
    `` \,\mathrm{\,Again\,},\,\mathrm{\,for\,}n=2,``
    `` {\lambda }_{2}={y}_{n}\frac{d}{2D}=500\,\mathrm{\,nm\,}``
    So, the light of wavelength 500 nm will have strong intensity.
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  • #25-b
    Which wavelength(s) will have a strong intensity?
    Figure
    Ans : The wavelength(s) will have a strong intensity, which will form a bright fringe at the position of the hole.
    `` \,\mathrm{\,So\,},{y}_{n}=n{\lambda }_{n}\frac{D}{d}``
    `` \Rightarrow {\lambda }_{n}={y}_{n}\frac{d}{nD}``
    `` \,\mathrm{\,For\,}n=1,``
    `` {\lambda }_{1}={y}_{n}\frac{d}{D}``
    `` ={10}^{-3}\times \left(0.5\right)\times \frac{{10}^{-3}}{0.5}``
    `` ={10}^{-6}\,\mathrm{\,m\,}=1000\,\mathrm{\,nm\,}.``
    But 1000 nm does not fall in the range 400 nm - 700 nm.
    `` \,\mathrm{\,Again\,},\,\mathrm{\,for\,}n=2,``
    `` {\lambda }_{2}={y}_{n}\frac{d}{2D}=500\,\mathrm{\,nm\,}``
    So, the light of wavelength 500 nm will have strong intensity.
    Page No 382: