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NEET-XII-Physics

17: Light Waves

with Solutions - page 3
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  • #5
    Two narrow slits emitting light in phase are separated by a distance of 1⋅0 cm. The wavelength of the light is
    5·0×10-7 m. The interference pattern is observed on a screen placed at a distance of 1⋅0 m. (a) Find the separation between consecutive maxima. Can you expect to distinguish between these maxima? (b) Find the separation between the sources which will give a separation of 1⋅0 mm between consecutive maxima. (a) Find the separation between consecutive maxima. Can you expect to distinguish between these maxima? (b) Find the separation between the sources which will give a separation of 1⋅0 mm between consecutive maxima.
    Ans : Given:
    Separation between two narrow slits, d = 1 cm = 10-2 m
    Wavelength of the light, `` \lambda =5\times {10}^{-7}\,\mathrm{\,m\,}``
    `` ``
    Distance of the screen, `` D=1\,\mathrm{\,m\,}`` (a) We know that separation between two consecutive maxima = fringe width (β).
    That is, `` \beta =\frac{\lambda D}{d}`` ...(i)
    `` =\frac{5\times {10}^{-7}\times 1}{{10}^{-2}}\,\mathrm{\,m\,}``
    `` =5\times {10}^{-5}\,\mathrm{\,m\,}=0.05\,\mathrm{\,mm\,}`` (b) Separation between two consecutive maxima = fringe width
    ∴ `` \beta =1\,\mathrm{\,mm\,}={10}^{-3}\,\mathrm{\,m\,}``
    Let the separation between the sources be 'd'
    Using equation (i), we get:
    `` d\text{'}=\frac{5\times {10}^{-7}\times 1}{{10}^{-3}}``
    `` \Rightarrow d\text{'}=5\times {10}^{-4}\,\mathrm{\,m\,}=0.50\,\mathrm{\,mm\,}.``
    Page No 380:
    Page No 381: (a) We know that separation between two consecutive maxima = fringe width (β).
    That is, `` \beta =\frac{\lambda D}{d}`` ...(i)
    `` =\frac{5\times {10}^{-7}\times 1}{{10}^{-2}}\,\mathrm{\,m\,}``
    `` =5\times {10}^{-5}\,\mathrm{\,m\,}=0.05\,\mathrm{\,mm\,}`` (b) Separation between two consecutive maxima = fringe width
    ∴ `` \beta =1\,\mathrm{\,mm\,}={10}^{-3}\,\mathrm{\,m\,}``
    Let the separation between the sources be 'd'
    Using equation (i), we get:
    `` d\text{'}=\frac{5\times {10}^{-7}\times 1}{{10}^{-3}}``
    `` \Rightarrow d\text{'}=5\times {10}^{-4}\,\mathrm{\,m\,}=0.50\,\mathrm{\,mm\,}.``
    Page No 380:
    Page No 381:
  • #5-a
    Find the separation between consecutive maxima. Can you expect to distinguish between these maxima?
    Ans : We know that separation between two consecutive maxima = fringe width (β).
    That is, `` \beta =\frac{\lambda D}{d}`` ...(i)
    `` =\frac{5\times {10}^{-7}\times 1}{{10}^{-2}}\,\mathrm{\,m\,}``
    `` =5\times {10}^{-5}\,\mathrm{\,m\,}=0.05\,\mathrm{\,mm\,}``
  • #5-b
    Find the separation between the sources which will give a separation of 1⋅0 mm between consecutive maxima.
    Ans : Separation between two consecutive maxima = fringe width
    ∴ `` \beta =1\,\mathrm{\,mm\,}={10}^{-3}\,\mathrm{\,m\,}``
    Let the separation between the sources be 'd'
    Using equation (i), we get:
    `` d\text{'}=\frac{5\times {10}^{-7}\times 1}{{10}^{-3}}``
    `` \Rightarrow d\text{'}=5\times {10}^{-4}\,\mathrm{\,m\,}=0.50\,\mathrm{\,mm\,}.``
    Page No 380:
    Page No 381: