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NEET-XII-Physics

17: Light Waves

with Solutions - page 4

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  • #2-a
    Find its frequency in air.
    Ans : Frequency in air, `` {f}_{\,\mathrm{\,air\,}}=\frac{c}{{\lambda }_{\,\mathrm{\,a\,}}}``
    `` {f}_{\,\mathrm{\,air\,}}=\frac{3\times {10}^{8}}{589\times {10}^{-9}}``
    `` =5.09\times {10}^{14}\,\mathrm{\,Hz\,}``
  • #2-b
    Find its wavelength in water (refractive index = 1.33).
    Ans : Let wavelength of sodium light in water be `` {\lambda }_{\,\mathrm{\,w\,}}``.
    We know that
    `` \frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,a\,}}}{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,w\,}}}=\frac{{\lambda }_{\,\mathrm{\,\omega \,}}}{{\lambda }_{\,\mathrm{\,a\,}}}``,
    where μa is the refractive index of air which is equal to 1 and
    λw is the wavelength of sodium light in water.
    `` \Rightarrow \frac{1}{1.33}=\frac{{\lambda }_{\,\mathrm{\,\omega \,}}}{589\times {10}^{-9}}``
    `` \Rightarrow {\lambda }_{\,\mathrm{\,\omega \,}}=443\,\mathrm{\,nm\,}``
  • #2-c
    Find its frequency in water.
    Ans : Frequency of light does not change when light travels from one medium to another.
    `` \therefore {f}_{\,\mathrm{\,\omega \,}}={f}_{\,\mathrm{\,a\,}}``
    `` =5.09\times {10}^{14}\,\mathrm{\,Hz\,}``
  • #2-d
    Find its speed in water.
    Ans : Let the speed of sodium light in water be `` {\nu }_{\,\mathrm{\,\omega \,}}``
    and speed in air, va = c.
    Using `` \frac{{\mu }_{\,\mathrm{\,a\,}}}{{\mu }_{\,\mathrm{\,\omega \,}}}=\frac{{\nu }_{\,\mathrm{\,\omega \,}}}{{\nu }_{\,\mathrm{\,a\,}}}``, we get:
    `` {\nu }_{\,\mathrm{\,\omega \,}}=\frac{{\mu }_{\,\mathrm{\,a\,}}c}{{\mu }_{\,\mathrm{\,\omega \,}}}``
    `` =\frac{3\times {10}^{8}}{\left(1.33\right)}=2.25\times {10}^{8}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
    Page No 380:
  • Qstn #3
    The index of refraction of fused quartz is 1â‹…472 for light of wavelength 400 nm and is 1â‹…452 for light of wavelength 760 nm. Find the speeds of light of these wavelengths in fused quartz.
    Ans : Given:
    Refractive index of fused quartz for light of wavelength 400 nm is 1.472.
    And refractive index of fused quartz for light of wavelength 760 nm is 1.452.
    We known that refractive index of a material is given by
    μ = `` \frac{\,\mathrm{\,Speed\,}\,\mathrm{\,of\,}\,\mathrm{\,light\,}\,\mathrm{\,in\,}\,\mathrm{\,vacuum\,}}{\,\mathrm{\,Speed\,}\,\mathrm{\,of\,}\,\mathrm{\,light\,}\,\mathrm{\,in\,}\,\mathrm{\,the\,}\,\mathrm{\,material\,}}=\frac{c}{v}``
    Let speed of light for wavelength 400 nm in quartz be v400.
    So,
    `` 1.472=\frac{3\times {10}^{8}}{{v}_{400}}``
    `` \Rightarrow {v}_{400}=2.04\times {10}^{8}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
    `` ``
    Let speed of light of wavelength 760 nm in quartz be v760.
    Again, `` \frac{1.452}{1}=\frac{3\times {10}^{8}}{{\nu }_{760}}``
    `` \Rightarrow {v}_{760}=2.07\times {10}^{8}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
    Page No 380:
  • Qstn #4
    The speed of yellow light in a certain liquid is 2⋅4 × 108 m s-1. Find the refractive index of the liquid.
    Ans : Given:
    Speed of yellow light in liquid (vL)= 2⋅4 × 108 m s-1
    And speed of yellow light in air speed = va
    Let μL be the refractive index of the liquid
    Using, `` {\,\mathrm{\,\mu \,}}_{\,\mathrm{\,L\,}}=\frac{\,\mathrm{\,speed\,}\,\mathrm{\,of\,}\,\mathrm{\,light\,}\,\mathrm{\,in\,}\,\mathrm{\,vaccum\,}}{\,\mathrm{\,velocity\,}\,\mathrm{\,of\,}\,\mathrm{\,light\,}\,\mathrm{\,in\,}\,\mathrm{\,the\,}\,\mathrm{\,given\,}\,\mathrm{\,medium\,}}=\frac{c}{{v}_{\,\mathrm{\,L\,}}}``
    `` {\,\mathrm{\,\mu \,}}_{\,\mathrm{\,L\,}}=\frac{3\times {10}^{8}}{\left(2.4\right)\times {10}^{8}}=1.25``
    `` ``
    Hence, the required refractive index is 1.25.
    Page No 380:
  • Qstn #5
    Two narrow slits emitting light in phase are separated by a distance of 1â‹…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.
    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\,}``
  • #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 1â‹…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:
  • Qstn #6
    The separation between the consecutive dark fringes in a Young’s double slit experiment is 1⋅0 mm. The screen is placed at a distance of 2⋅5 m from the slits and the separation between the slits is 1⋅0 mm. Calculate the wavelength of light used for the experiment.
    Ans : Given:
    Separation between consecutive dark fringes = fringe width (β) = 1 mm = 10-3 m
    Distance between screen and slit (D) = 2.5 m
    The separation between slits (d) = 1 mm = 10-3 m
    Let the wavelength of the light used in experiment be λ.
    We know that
    `` \beta =\frac{\lambda D}{d}``
    `` {10}^{-3}\,\mathrm{\,m\,}=\frac{2.5\times \lambda }{{10}^{-3}}``
    `` \Rightarrow \lambda =\frac{1}{2.5}{10}^{-6}\,\mathrm{\,m\,}``
    `` =4\times {10}^{-7}\,\mathrm{\,m\,}=400\,\mathrm{\,nm\,}``
    Hence, the wavelength of light used for the experiment is 400 nm.
    Page No 381:
  • Qstn #7
    In a double slit interference experiment, the separation between the slits is 1⋅0 mm, the wavelength of light used is 5⋅0 × 10-7 m and the distance of the screen from the slits is 1⋅0 m.
    Ans : Given:
    Separation between the two slits, `` d=1\,\mathrm{\,mm\,}={10}^{-3}\,\mathrm{\,m\,}``
    Wavelength of the light used, `` \lambda =5.0\times {10}^{-7}\,\mathrm{\,m\,}``
    `` ``
    Distance between screen and slit, `` D=1\,\mathrm{\,m\,}``
  • #7-a
    Find the distance of the centre of the first minimum from the centre of the central maximum.
    Ans : The distance of the centre of the first minimum from the centre of the central maximum, x = `` \frac{\,\mathrm{\,width\,}\,\mathrm{\,of\,}\,\mathrm{\,central\,}\,\mathrm{\,maxima\,}}{2}``
    That is, `` x=\frac{\beta }{2}=\frac{\lambda D}{2d}`` ...(i)
    `` =\frac{5\times {10}^{-7}\times 1}{2\times {10}^{-3}}``
    `` =2.5\times {10}^{-4}\,\mathrm{\,m\,}=0.25\,\mathrm{\,mm\,}``
  • #7-b
    How many bright fringes are formed in one centimetre width on the screen?
    Ans : From equation (i),
    fringe width, `` \beta =2\times x=0.50\,\mathrm{\,mm\,}``
    So, number of bright fringes formed in one centimetre (10 mm) = `` \frac{10}{0.50}=20``.
    Page No 381:
  • Qstn #8
    In a Young’s double slit experiment, two narrow vertical slits placed 0⋅800 mm apart are illuminated by the same source of yellow light of wavelength 589 nm. How far are the adjacent bright bands in the interference pattern observed on a screen 2⋅00 m away?
    Ans : Given:
    Separation between two narrow slits, `` d=0.8\,\mathrm{\,mm\,}=0.8\times {10}^{-3}\,\mathrm{\,m\,}``
    Wavelength of the yellow light, `` \lambda =589\,\mathrm{\,nm\,}=589\times {10}^{-9}\,\mathrm{\,m\,}``
    `` ``
    Distance between screen and slit, `` D=2.0\,\mathrm{\,m\,}``
    Separation between the adjacent bright bands = width of one dark fringe
    That is, `` \beta =\frac{\lambda D}{d}`` ...(i)
    `` \Rightarrow \beta =\frac{589\times {10}^{-9}\times 2}{0.8\times {10}^{-3}}``
    `` =1.47\times {10}^{-3}\,\mathrm{\,m\,}``
    `` =1.47\,\mathrm{\,mm\,}.``
    Hence, the adjacent bright bands in the interference pattern are 1.47 mm apart.
    Page No 381:
  • Qstn #9
    Find the angular separation between the consecutive bright fringes in a Young’s double slit experiment with blue-green light of wavelength 500 nm. The separation between the slits is
    2·0×10-3 m.
    Ans : Given:
    Wavelength of the blue-green light, `` \lambda =500\times {10}^{-9}\,\mathrm{\,m\,}``
    `` ``
    Separation between two slits, `` d=2\times {10}^{-3}\,\mathrm{\,m\,},``
    Let angular separation between the consecutive bright fringes be θ.
    `` \,\mathrm{\,Using\,}\theta =\frac{\beta }{D}=\frac{\lambda D}{dD}=\frac{\lambda }{d},\text{we get:}``
    `` \theta =\frac{500\times {10}^{-9}}{2\times {10}^{-3}}``
    `` =250\times {10}^{-6}``
    `` =25\times {10}^{-5}\,\mathrm{\,radian\,}\,\mathrm{\,or\,}0.014°``
    Hence, the angular separation between the consecutive bright fringes is 0.014 degree.
    Page No 381:
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