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Wave Optics Test - 5
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Wave Optics Test - 5
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  • Question 1/10
    1 / -0.25

    Is light a particle or a wave?

    Solutions

    Explanation:explanation none

  • Question 2/10
    1 / -0.25

    Referring to the Young ’s double slit experiment, Phase difference corresponding to a Path Difference of  λ /3 is

    Solutions

    We know that,
    Phase difference=K ×path difference. K=2 π/ λ.
    Therefore, phase difference=(2 π/ λ) ×( λ/3)=2 π/3=120 degree.

  • Question 3/10
    1 / -0.25

    The angle of incidence at which the reflected beam is fully polarized is called

    Solutions

    Brewster 's angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. When unpolarized light is incident at this angle, the light that is reflected from the surface is therefore perfectly polarized.

  • Question 4/10
    1 / -0.25

    The refractive index of glass is 1.5 which of the two colors red and violet travels slower in a glass prism?

    Solutions

    (a) Speed of light in glass is  c/μ=2 ×108 m/s

    (b) The speed of light in glass is not independent of the colour of light. The refractive index of a violet component of white light is greater than the refractive  index of a red component. Hence, the speed of violet light is less than the speed of red  light in glass. Hence, violet light travels slower than red light in a glass prism.

  • Question 5/10
    1 / -0.25

    In a young ’s double slit experiment, the central bright fringe can be identified by

    Solutions

    When we use white light, the central bright will have light from all wavelengths as none of them cancel out. Hence the central bright fringe appears white.
    For other bright fringes, depending on the wavelength of light constructive interference will not take place for certain wavelengths. Hence they will not be white, rather will be coloured, hence differentiated from central fringe.

  • Question 6/10
    1 / -0.25

    The propagation of light is best described by,

    Solutions

    Light as a wave: Light can be described (modeled) as an electromagnetic wave. In this model, a changing electric field creates a changing magnetic field. This changing magnetic field then creates a changing electric field and BOOM - you have light. Unlike many other waves (sound, water waves, waves in a football stadium), light does not need a medium to “wave ”in.

  • Question 7/10
    1 / -0.25

    In general the term diffraction is used

    Solutions

    The process by which a beam of light or other system of waves is spread out as a result of passing through a narrow aperture or across an edge, typically accompanied by interference between the waveforms produced.

  • Question 8/10
    1 / -0.25

    Shape of the wave front of light diverging from a point source is

    Solutions

    The shape of the wave front in case of a light diverging from a point source is spherical and is shown in the given figure. The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid.

  • Question 9/10
    1 / -0.25

    Emission and absorption is best described by,

    Solutions

    The Particle Model of emission and absorption is based on the idea that light is composed of discrete particles called photons , and these photons carry quantized energy. This model helps us understand how atoms and molecules interact with light by either emitting or absorbing photons during electronic transitions.

  • Question 10/10
    1 / -0.25

    In a Young ’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment

    Solutions

    Distance between the slits, d=0.28 ×10 −3  m

    Distance between the slits and the screen, D=1.4m

    Distance between the central fringe and the fourth  (n=4) fringe, 

    u=1.2 ×10 −2  m

    In case of a constructive interference, we have the relation for the distance between the two fringes as:

    u=n λD/d

    ⇒λ=ud/nD=6 ×10 −7m=600nm

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