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NEET-XII-Physics

17: Light Waves

with Solutions - page 6

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  • Qstn #19
    A Young’s double slit apparatus has slits separated by 0⋅28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by red light
    λ=700 nm in vacuum. Find the fringe-width of the pattern formed on the screen.
    Ans : Given:
    Separation between two slits, `` d=0.28\,\mathrm{\,mm\,}=0.28\times {10}^{-3}\,\mathrm{\,m\,}``
    `` ``
    Distance between screen and slit (D) = 48 cm = 0.48 m
    Wavelength of the red light, `` {\lambda }_{a}=700\,\mathrm{\,nm\,}\,\mathrm{\,in\,}\,\mathrm{\,vaccum\,}=700\times {10}^{-9}\,\mathrm{\,m\,}``
    Let the wavelength of red light in water = `` {\lambda }_{\omega }``
    We known that refractive index of water (μw =4/3),
    μw = `` \frac{\,\mathrm{\,Speed\,}\,\mathrm{\,of\,}\,\mathrm{\,light\,}\,\mathrm{\,in\,}\,\mathrm{\,vacuum\,}}{\,\mathrm{\,Speed\,}\,\mathrm{\,of\,}\,\mathrm{\,light\,}\,\mathrm{\,in\,}\,\mathrm{\,the\,}\,\mathrm{\,water\,}}``
    `` \,\mathrm{\,So\,},{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,w\,}}=\frac{{v}_{a}}{{v}_{\omega }}=\frac{{\lambda }_{a}}{{\lambda }_{\omega }}``
    `` \Rightarrow \frac{4}{3}=\frac{{\lambda }_{a}}{{\lambda }_{\omega }}``
    `` \Rightarrow {\lambda }_{\omega }=\frac{3{\lambda }_{a}}{4}=\frac{3\times 700}{4}=525\,\mathrm{\,nm\,}``
    So, the fringe width of the pattern is given by
    `` \,\mathrm{\,\beta \,}=\frac{{\lambda }_{\omega }D}{d}``
    `` =\frac{525\times {10}^{-9}\times \left(0.48\right)}{\left(0.28\right)\times {10}^{-3}}``
    `` =9\times {10}^{-4}=0.90\,\mathrm{\,mm\,}``
    Hence, fringe-width of the pattern formed on the screen is 0.90 mm.
    Page No 381:
  • Qstn #20
    A parallel beam of monochromatic light is used in a Young’s double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Slow that if the incident beam makes an angle
    θ=sin-1 λ2dwith the normal to the plane of the slits, there will be a dark fringe at the centre P0 of the pattern.
    Ans : Let the two slits are S1 and S2 with separation d as shown in figure.

    The wave fronts reaching P0 from S1 and S2 will have a path difference of S1X = ∆x.
    In ∆S1S2X, `` \,\mathrm{\,sin\,}\theta =\frac{{\,\mathrm{\,S\,}}_{1}\,\mathrm{\,X\,}}{{\,\mathrm{\,S\,}}_{1}{\,\mathrm{\,S\,}}_{2}}=\frac{∆x}{d}``
    `` \Rightarrow ∆x=d\,\mathrm{\,sin\,}\theta ``
    `` ``
    Using `` \,\mathrm{\,\theta \,}={\,\mathrm{\,sin\,}}^{-1}\left(\frac{\lambda }{2d}\right)``, we get,
    `` \Rightarrow ∆x=d\times \frac{\lambda }{2d}=\frac{\lambda }{2}``
    Hence, there will be dark fringe at point P0 as the path difference is an odd multiple of `` \frac{\lambda }{2}``.
    Page No 381:
  • Qstn #21
    A narrow slit S transmitting light of wavelength
    λis placed a distance d above a large plane mirror, as shown in the figure (17-E1). The light coming directly from the slit and that coming after the reflection interfere at a screen ∑ placed at a distance D from the slit.
  • #21-a
    What will be the intensity at a point just above the mirror, i.e. just above O?
    Ans : The phase of a light wave reflecting from a surface differs by '`` \pi ``' from the light directly coming from the source.

    Thus, the wave fronts reaching just above the mirror directly from the source and after reflecting from the mirror have a phase difference of `` \pi ``, which is the condition of distractive interference. So, the intensity at a point just above the mirror is zero.
  • #21-b
    At what distance from O does the first maximum occur?
    Figure
    Ans : Here, separation between two slits is `` 2d``.
    Wavelength of the light is `` \lambda ``.
    Distance of the screen from the slit is `` D``.
    Consider that the bright fringe is formed at position y. Then,
    path difference, `` ∆x=\frac{y\times 2d}{D}=n\lambda ``.
    After reflection from the mirror, path difference between two waves is `` \frac{\lambda }{2}``.
    `` \Rightarrow \frac{y\times 2d}{D}=\frac{\lambda }{2}+n\lambda ``
    `` \,\mathrm{\,For\,}\,\mathrm{\,first\,}or\,\mathrm{\,der\,},\,\mathrm{\,put\,}n=0.``
    `` \Rightarrow y=\frac{\lambda D}{4d}``
    Page No 382:
  • Qstn #22
    A long narrow horizontal slit is paced 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen 1â‹…0 m away from the slit. Find the fringe-width if the light used has a wavelength of 700 nm.
    Ans : Given:
    Separation between two slits, `` d\text{'}=2d=2\,\mathrm{\,mm\,}=2\times {10}^{-3}\,\mathrm{\,m\,}`` (as d = 1 mm)
    Wavelength of the light used, `` \lambda =700\,\mathrm{\,nm\,}=700\times {10}^{-3}\,\mathrm{\,m\,}``
    `` ``
    Distance between the screen and slit (D) = 1.0 m
    It is a case of Lloyd's mirror experiment.
    `` \,\mathrm{\,Fringe\,}\,\mathrm{\,width\,},\beta =\frac{\,\mathrm{\,\lambda D\,}}{d\text{'}}``
    `` =\frac{700\times {10}^{-9}\times 1}{2\times {10}^{-3}}``
    `` =0.35\times {10}^{-3}\,\mathrm{\,m\,}=0.35\,\mathrm{\,mm\,}``
    Hence, the width of the fringe is 0.35 mm.
    Page No 382:
  • Qstn #23
    Consider the situation of the previous problem. If the mirror reflects only 64% of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen?
    Ans : Given:
    The mirror reflects 64% of the energy or intensity of light.
    Let intensity of source = I1.
    And intensity of light after reflection from the mirror = I2.
    Let a1 and a2 be corresponding amplitudes of intensity I1 and I2.
    According to the question,
    `` {I}_{2}=\frac{{I}_{1}\times 64}{100}``
    `` \Rightarrow \frac{{I}_{2}}{{I}_{1}}=\frac{64}{100}=\frac{16}{25}``
    `` \,\mathrm{\,And\,}\frac{{I}_{2}}{{I}_{1}}=\frac{{{a}_{2}}^{2}}{{{a}_{1}}^{2}}``
    `` \Rightarrow \frac{{a}_{2}}{{a}_{1}}=\frac{4}{5}``
    `` \,\mathrm{\,We\,}\,\mathrm{\,know\,}\text{that}\frac{{I}_{\,\mathrm{\,max\,}}}{{I}_{\,\mathrm{\,min\,}}}=\frac{{\left({a}_{1}+{a}_{2}\right)}^{2}}{{\left({a}_{1}-{a}_{2}\right)}^{2}}``
    `` =\frac{{\left(5+4\right)}^{2}}{{\left(5-4\right)}^{2}}``
    `` {I}_{\,\mathrm{\,max\,}}:{I}_{\,\mathrm{\,min\,}}=81:1``
    Hence, the required ratio is 81 : 1.
    Page No 382:
  • Qstn #24
    A double slit S1 - S2 is illuminated by a coherent light of wavelength
    λ. The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance D1 from it and a screen ∑ is placed behind the double slit at a distance D2 from it (figure 17-E2). The screen ∑ receives only the light reflected by the mirror. Find the fringe-width of the interference pattern on the screen.
    Figure
    Ans : Given:
    Separation between the two slits = d
    Wavelength of the coherent light =λ
    Distance between the slit and mirror is D1.
    Distance between the slit and screen is D2.
    Therefore,
    apparent distance of the screen from the slits,
    `` D=2{D}_{1}+{D}_{2}``
    `` \,\mathrm{\,Fringe\,}\,\mathrm{\,width\,},\beta =\frac{\lambda D}{d}=\frac{\left(2{D}_{1}+{D}_{2}\right)\lambda }{d}``
    `` ``
    Hence, the required fringe width is `` \frac{\left(2{D}_{1}+{D}_{2}\right)\lambda }{d}``.
    Page No 382:
  • Qstn #25
    White coherent light (400 nm-700 nm) is sent through the slits of a Young’s double slit experiment (figure 17-E3). The separation between the slits is 0⋅5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1⋅0 mm away (along the width of the fringes) from the central line.
    Ans : Given: Separation between two slits,
    `` d=0.5\,\mathrm{\,mm\,}=0.5\times {10}^{-3}\,\mathrm{\,m\,}``
    `` ``
    Wavelength of the light, `` \lambda =400\,\mathrm{\,nm\,}\,\mathrm{\,to\,}700\,\mathrm{\,nm\,}``
    Distance of the screen from the slit, `` D=50\,\mathrm{\,cm\,}=0.5\,\mathrm{\,m\,}``,
    Position of hole on the screen, `` {y}_{n}=1\,\mathrm{\,mm\,}=1\times {10}^{-3}\,\mathrm{\,m\,}``
  • #25-a
    Which wavelength(s) will be absent in the light coming from the hole?
    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.
  • #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:
  • Qstn #26
    Consider the arrangement shown in the figure (17-E4). The distance D is large compared to the separation d between the slits. (a) Find the minimum value of d so that there is a dark fringe at O. (b) Suppose d has this value. Find the distance x at which the next bright fringe is formed. (c) Find the fringe-width.
    Figure
    digAnsr:   i
    Ans :
    From the figure, `` \,\mathrm{\,AB\,}=\,\mathrm{\,BO\,}\,\mathrm{\,and\,}\,\mathrm{\,AC\,}=\,\mathrm{\,CO\,}``.
    Path difference of the wave front reaching O,
    `` ∆x=\left(\,\mathrm{\,AB\,}+\,\mathrm{\,BO\,}\right)-\left(\,\mathrm{\,AC\,}+\,\mathrm{\,CO\,}\right)``
    `` =2\left(\,\mathrm{\,AB\,}-\,\mathrm{\,AC\,}\right)``
    `` =2\left(\sqrt{{D}^{2}+{d}^{2}}-D\right)``
    For dark fringe to be formed at O, path difference should be an odd multiple of `` \frac{\lambda }{2}``.
    `` \,\mathrm{\,So\,},∆x=\left(2n+1\right)\frac{\lambda }{2}``
    `` \Rightarrow 2\left(\sqrt{{D}^{2}+{d}^{2}}-D\right)=\left(2n+1\right)\frac{\lambda }{2}``
    `` \Rightarrow \sqrt{{D}^{2}+{d}^{2}}=D+\left(2n+1\right)\frac{\lambda }{4}``
    `` \Rightarrow {D}^{2}+{d}^{2}={D}^{2}+{\left(2n+1\right)}^{2}\frac{{\lambda }^{2}}{16}+\left(2n+1\right)\frac{\lambda D}{2}``
    Neglecting, `` {\left(2n+1\right)}^{2}\frac{{\lambda }^{2}}{16}``
    `` ``, as it is very small, we get:
    `` d=\left(\sqrt{2n}+1\right)\frac{\lambda D}{2}``
    `` \,\mathrm{\,For\,}\,\mathrm{\,minimum\,}d,\,\mathrm{\,putting\,},n=0``
    `` {d}_{\,\mathrm{\,min\,}}=\sqrt{\left(\frac{\lambda D}{2}\right)}``, we get:
    Thus, for `` {d}_{\,\mathrm{\,min\,}}=\sqrt{\left(\frac{\lambda D}{2}\right)}`` there is a dark fringe at O.
    Page No 382:
  • Qstn #27
    Two coherent point sources S1 and S2, vibrating in phase, emit light of wavelength
    λ. The separation between the sources is
    2λ. Consider a line passing through S2 and perpendicular to the line S1 S2. What is the smallest distance from S2 where a minimum intensity occurs?
    Ans :
    Let P be the point of minimum intensity.
    For minimum intensity at point P,
    `` {S}_{1}\,\mathrm{\,P\,}-{S}_{2}\,\mathrm{\,P\,}=x=\left(2n+1\right)\frac{\lambda }{2}``
    Thus, we get:
    `` \sqrt{{\,\mathrm{\,Z\,}}^{2}+{\left(2\lambda \right)}^{2}}-\,\mathrm{\,Z\,}=\left(2n+1\right)\frac{\lambda }{2}``
    `` \Rightarrow {\,\mathrm{\,Z\,}}^{2}+4{\lambda }^{2}={\,\mathrm{\,Z\,}}^{2}{\left(2n+1\right)}^{2}\frac{{\lambda }^{2}}{4}+2\,\mathrm{\,Z\,}\left(2n+1\right)\frac{\lambda }{2}``
    `` \Rightarrow \,\mathrm{\,Z\,}=\frac{4{\lambda }^{2}-\left(2n+1\right)2{\lambda }^{2}/4}{\left(2n+1\right)\lambda }``
    `` =\frac{16{\lambda }^{2}-\left(2n+1\right)}{4\left(2n+1\right)\lambda }...\left(i\right)``
    `` \text{When}n=0,\text{Z}=\frac{15\lambda }{4}``
    `` n=-1,\,\mathrm{\,Z\,}=\frac{-15\lambda }{4}``
    `` n=1,\,\mathrm{\,Z\,}=\frac{7\lambda }{12}``
    `` n=2,\text{Z}=\frac{-9\lambda }{20}``
    Thus, `` \,\mathrm{\,Z\,}=\frac{7\lambda }{12}`` is the smallest distance for which there will be minimum intensity.
    Page No 382:
  • Qstn #28
    Figure (17-E5) shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let
    BP0-AP0=λ/3 and D>>λ.
  • #28-a
    Show that in this case
    d=2λD/3.
    Ans : Given:
    Wavelength of light = `` \lambda ``
    Path difference of wave fronts reaching from A and B is given by
    `` ∆{x}_{B}={\,\mathrm{\,BP\,}}_{0}-{\,\mathrm{\,AP\,}}_{0}=\frac{\lambda }{3}``
    `` \Rightarrow \sqrt{{D}^{2}+{d}^{2}}-D=\frac{\lambda }{3}``
    `` \Rightarrow {D}^{2}+{d}^{2}={D}^{2}+\frac{{\lambda }^{2}}{9}+\frac{2\lambda D}{3}``
    We will neglect the term `` \frac{{\lambda }^{2}}{9}``, as it has a very small value.
    `` \therefore d=\sqrt{\frac{\left(2\lambda D\right)}{3}}``
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