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    Ruba Presentation

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    Shifa Naseer
    Total internal reflection occurs when light travels from an optically dense medium to a less dense medium at an angle greater than the critical angle. The critical angle is the maximum angle of incidence above which total internal reflection occurs. Total internal reflection results in no light being transmitted into the second medium and all light being reflected back into the first medium. Applications of total internal reflection include light pipes and endoscopes. Snell's law and the refractive indices of the media can be used to calculate the critical angle.

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    Ruba Presentation

    Uploaded by

    Shifa Naseer
    21 views12 pages
    Total internal reflection occurs when light travels from an optically dense medium to a less dense medium at an angle greater than the critical angle. The critical angle is the maximum angle of incidence above which total internal reflection occurs. Total internal reflection results in no light being transmitted into the second medium and all light being reflected back into the first medium. Applications of total internal reflection include light pipes and endoscopes. Snell's law and the refractive indices of the media can be used to calculate the critical angle.

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    Total internal reflection occurs when light travels from an optically dense medium to a less dense medium at an angle greater than the critical angle. The critical angle is the maximum angle of incidence above which total internal reflection occurs. Total internal reflection results in no light being transmitted into the second medium and all light being reflected back into the first medium. Applications of total internal reflection include light pipes and endoscopes. Snell's law and the refractive indices of the media can be used to calculate the critical angle.

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    21 views12 pages

    Ruba Presentation

    Uploaded by

    Shifa Naseer
    Total internal reflection occurs when light travels from an optically dense medium to a less dense medium at an angle greater than the critical angle. The critical angle is the maximum angle of incidence above which total internal reflection occurs. Total internal reflection results in no light being transmitted into the second medium and all light being reflected back into the first medium. Applications of total internal reflection include light pipes and endoscopes. Snell's law and the refractive indices of the media can be used to calculate the critical angle.

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    TOTAL INTERNAL

    REFLECTION
    RUBA AND TALIQA
    REFLECTION AND REFRACTION
    • Reflection involves a Refraction of waves involves a
    change in direction of change in the direction of
    waves when they waves as they pass from one
    bounce off a barrier.  medium to another. 
    Refraction, or the bending of
    the path of the waves, is
    accompanied by a change in
    speed and wavelength of the
    waves.
    TOTAL INTERNAL REFLECTION
    CRITICAL ANGLE
    The angle of incidence
    beyond which rays of
    light passing through a
    denser medium to the
    surface of a less dense
    medium are no longer
    refracted but totally
    reflected.
    TOTAL INTERNAL REFLECTION
    Total internal reflection, in
    physics,
    complete reflection of a ray
    of light within a medium such
    as water or glass from the
    surrounding surfaces back
    into the medium. The
    phenomenon occurs if the
    angle of incidence is greater
    than a certain limiting angle,
    called the critical angle
    CONDITIONS FOR TOTAL INTERNAL
    REFLECTION
    For total internal reflection to occur,
    2 conditions must be met:
    • Ray of light must be travel from
    denser medium to rare medium.
    • The angle of incidence of the light
    ray must exceed the critical angle
    of the interface.
    TOTAL INTERNAL REFLECTION
    APPLICTIONS
    • Light Pipe:
    It is bundle of thousands of optical fibers bounded
    together. They are used to illuminate the inaccessible places by the
    doctors or engineers.
    • Endoscope:
    It is a medical instrument used for exploratory
    diagnostic and surgical purposes.
    Refractive index
    the refractive index or index of refraction (n) of
    a material is a dimensionless number that
    describes how fast light propagates through the
    material. It is defined as

    n=

    where c is the speed of light in vacuum and v is


    the phase velocity of light in the medium. For
    example, the refractive index of water is 1.333,
    meaning that light travels 1.333 times as fast in
    vacuum as in water. The refractive index
    determines how much the path of light is bent, or
    refracted, when entering a material.
    HOW TO FIND A CRITICAL ANGLE
    n1 sinθ1 = n2 sinθ2 (snells law)

    (1.5) sinθC = (1.0) sin 90

    sinθc =
    θc =

    θc = 41.8˚
    QUESTIONS
    A scuba diver is wearing a head lamp and looking up at the surface of the water.
    If the minimum angle to the vertical resulting in total internal reflection is 25∘,
    what is the index of refraction of the water?
    θ air =1.00 HINT(SNELLS LAW= n1 sinθ1 = n2 sinθ2 ANS=2.37

    Explanation:
    We can use Snell's law:
    n1sin(θ1)=n2sin(θ2)

    At the first instance of total internal reflection, the angle of light on the air side will
    be 90∘ to the vertical, so the equation becomes:

    n1sin(θ1)=n2
    Rearranging for the index of refraction of water, we get:

    n1=n2sin(θ1)=1.0sin(25∘)=2.37
    QUESTION
    You have a piece of optical wire with a light beam traveling through it
    surrounded by a liquid with an index of refraction of 1.33. If the index of
    refraction of the wire is 1.85, what is the critical angle needed to achieve
    total internal reflection?
    Explanation:
    The equation to find total internal reflection is a special case of Snell's Law, with the
    refraction angle set to 90˚. This new equation looks like this:
    =

    To find the critical angle, we can rearrange the equation.

    As you can tell from this, the index of refraction of the medium the light is traveling
    from must be higher than the index of refraction of the other medium. Our numbers
    qualify. Now, we can plug in our values to find the critical angle.                      
    =(1.331.85)
    =45.97∘

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