A ray of light strikes the midpoint of one face of an equiangular (60°–60°–60°) glass prism (n = 1.5) at an angle of incidence of 39.8°. (a)

Question

A ray of light strikes the midpoint of one face of an equiangular (60°–60°–60°) glass prism (n = 1.5) at an angle of incidence of 39.8°. (a) Trace the path of the light ray through the glass, and find the angles of incidence and refraction at each surface.

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Nho 6 months 2021-07-22T03:15:13+00:00 1 Answers 17 views 0

Answers ( )

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    2021-07-22T03:16:59+00:00

    Answer:

    a

    The path of light ray through the glass is shown on the first uploaded image

    First surface:

    Angle of  incidence is  i_1 = 39.0^o

    Angle of  refraction  is  r_1= 25.23^o

    Second surface:

    Angle of  incidence is  i_2 = 34.77^o

    Angle of  refraction  is  r_2= 58.80^o

    b

    Since the angle of incidence is equal to the angle of reflection

    Then at the first surface the angle of reflection  is  R = 39.8^o

    And at the first surface the angle of reflection  is  R_2 = 34.77^o

    Explanation:

    From the question we are told that

         The angle of incidence is i_1 = 39.0^o

          The refractive index of the prism is n_{p} = 1.5

          The angle of the prism is A = 60^o

    The path of light ray through the glass is shown on the first uploaded image

      For the first surface of the prism

            According to Snell’s law

                     n_{air} sin( i_1) = n_{p} sin( r_1)

    The refractive index of air n_{air}  has a constant value of  1

    Now making  the angle of refraction at the first surface of the prism r_1 the subject

                 r_1 = sin^{-1}[\frac{sin (i_1) }{n_{p}} ]

                    = sin^{-1}[\frac{sin(39.8)}{1.5} ]

                   r_1= 25.23^o

      For the second  surface of the prism

     looking at the diagram on the first uploaded image the angle of incidence is mathematically evaluated  as

                        i_2 = A - r_1

    Substituting values

                        i_2 = 60 -25.23

                             =34.77^o

            According to Snell’s law

                     n_{p} sin( i_2) = n_{air} sin( r_2)

    Now making  the angle of refraction at the second surface of the prism r_2 the subject

                    r_2 = sin^{-1} [\frac{n_p sin(i_2)}{n_air} ]

    Substituting values into the equation

                    r_2 = sin^{-1} [\frac{1.5 * sin(34.77)}{1}]

                   r_2 =58.8^o

     

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