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Wave Optics Quiz 9

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JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get a detailed list of all topics that are important in cracking the entrance examination. JEE Advanced syllabus for Physics has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with day-to-day experiences. In comparison to the other two subjects, the syllabus of JEE Advanced for physics is developed in such a way so as to test the deep understanding and application of concepts. .

Q1. In Young’s double-slit experiment, the angular width of a fringe formed on a distant screen is 1°. The wavelength of light used is 6000 Å. What is the spacing between the slits?
  •  344 mm
  •  0.1344 mm
  •  0.0344 mm
  •  0.034 mm
Solution

Q2.In Young’s double-slit experiment, the intensity of light at a point on the screen where path difference is 𝜆 is 𝐞. If intensity at a point is 𝐞/4, then possible path difference at this point are
  •  ðœ†/2, 𝜆/3
  •  ðœ†/3, 2𝜆/3
  •  ðœ†/3, 𝜆/4
  •  2𝜆/3, 𝜆/4
Solution
Q3.  In a Young’s double-slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen is given by
  •  12
  •  18
  •  24
  •  30
Solution
Q4. In Young’s double-slit interference experiment, if the slit separation is made threefold, the fringe width becomes
  •  Sixfold
  •  Threefold
  •  3/6-fold
  •  1/3-fold
Solution

Q5.Light of wavelength 𝜆0 in air enters a medium of refractive index 𝑛. If two points ðī and ðĩ in this medium lie along the path of this light at a distance ð‘Ĩ, then phase difference 𝜙0 between these two points is
  •   𝜙0=
    1 / n
    (
    2𝜋 / 𝜆0
    )x
  •   𝜙0=n(
    2𝜋 / 𝜆0
    )x
  •   𝜙0=(n-1)(
    2𝜋 / 𝜆0
    )x
  •   𝜙0=
    1 / (n-1)
    (
    2𝜋 / 𝜆0
    )x
Solution

Q6. Two slits spaced 0.25 mm apart are placed 0.75 m from a screen and illuminated by coherent light with a wavelength of 650 nm. The intensity at the center of the central maximum (𝜃 = 0°) is 𝐞0. The distance on the screen from the center of the central maximum to the point where the intensity has fallen to 𝐞0/2 is nearly
  •  0.1mm
  •  .25mm
  • 0.4mm
  •  0.5mm
Solution
Q7.A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of incident beam. At the first maxima of the diffraction pattern the phase difference between the rays coming from the edges of the slit is
  •  0
  •  ðœ‹/2
  •  ðœ‹
  •  2𝜋
Solution
Q8.In YDSE, if a bichromatic light having wavelengths 𝜆1 and 𝜆2 is used, then the maxima due to both lights will overlap at a certain distance ð‘Ķ of from the central maxima. Take separation between slits as 𝑑 and distance between screen and slits as 𝐷. Then, the value of ð‘Ķ will be
  •  (
    𝜆1+𝜆2 / 2D
    )d
  •  (
    𝜆1-𝜆2 / D
    )2d
  •  LCM of
    𝜆1D / d
    and
    𝜆2D / d
  •  HCF of
    𝜆1D / d
    and
    𝜆2D / d
Solution
Q9.In figure, if a parallel beam of while light is incident on the plane of the slits, then the distance of the nearest white spot on the screen from 𝑂 is [assume 𝑑 ≪ 𝐷, 𝜆 ≪ 𝑑]
  •  0
  •  ð‘‘/2
  •  ð‘‘/3
  •  ð‘‘/6
Solution
Q10. In Young’s double-slit experiment using monochromatic light of wavelength 𝜆, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 2.0 Ξm is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the plane of slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of the light
  •  4500Å
  •  5700Å
  •  6000Å
  • 4000Å
Solution

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