A laser beam enters a 16.5 cm thick glass window at an angle of 58.0° from the normal. The index of refraction of the glass is 1.47. At what

Question

A laser beam enters a 16.5 cm thick glass window at an angle of 58.0° from the normal. The index of refraction of the glass is 1.47. At what angle from the normal does the beam travel through the glass?

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Nem 4 years 2021-08-05T17:42:38+00:00 2 Answers 28 views 0

Answers ( )

  1. thachthao
    0
    2021-08-05T17:43:40+00:00
    Reply

    Answer:

    using Snells law

    Oi = angle of incidence = 58.0°

        ni = index of refraction of air = 1.0003

        nr = index of refraction of glass = 1.47

        c = speed of light in vacuum = 3 x 10^8 m/s

        Or = angle of refraction = ?

    ni(sinOi) = nR (sinOr)

    ni( sinOi)/ nR = sinOr

    arcsin(ni(sin0i))/nR = Or

    arcsin( 1.0003(sin58.0)) / 1.47

    Or = 35.25°

    Explanation:

  2. Euphemia
    0
    2021-08-05T17:43:57+00:00
    Reply

    Answer:

    = 35.23°

    time taken = 0.9898 ns

    Explanation:

    Using Snell’s law as:

    n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}

    Where,  

     is the angle of incidence  ( 58.0° )

    is the angle of refraction  ( ? )

    is the refractive index of the refraction medium  (glass, n=1.47)

    is the refractive index of the incidence medium (air, n=1)

    Hence,

    1\times {sin58.0^0}={1.47}\times{sin\theta_r}

    Angle of refraction= sin⁻¹ 0.5769 = 35.23°

    The distance it has to travel

    d =  = 16.5 cm / cos 35.23° = 20.20cm

    Also,

    n = c/v.

    Speed of light in vacuum = 3×10¹⁰ cm/s

    Speed in the medium is:

    v = c/n = 3×10¹⁰ cm/s / 1.47

      = 2.0408 × 10¹⁰ cm/s

    The time taken is:

    t = d/s

      = 20.20 cm / 2.0408×10¹⁰ cm/s

      = 9.898 × 10⁻¹⁰ s

       = 0.9898 × 10⁻⁹ s

    Also,

    1 ns = 10⁻⁹ s

    So, time taken = 0.9898 ns

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