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slega [8]
3 years ago
7

The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of

the light passing through it. At wavelength 486.1 nm (blue, designated with letter F) -> nF=1.497, and at wavelength 656.3 nm (red, designated with letter C) -> nC=1.488. Two beams (one of each wavelength) are prepared to coincide, and enter the flat polished surface of an acrylic block at angle of 45 arc degree measured from the normal to the surface. What is the angle between the blue beam and the red beam in the acrylic block?
Physics
1 answer:
Novosadov [1.4K]3 years ago
4 0

Answer:

The angle between the blue beam and the red beam in the acrylic block is  

 \theta _d  =0.19 ^o

Explanation:

From the question we are told that

     The  refractive index of the transparent acrylic plastic for blue light is  n_F  =  1.497

     The  wavelength of the blue light is F  =  486.1 nm  =  486.1 *10^{-9} \ m

    The  refractive index of the transparent acrylic plastic for red light is  n_C  =  1.488

       The  wavelength of the red light is C =  656.3 nm  = 656.3 *10^{-9} \  m

    The incidence angle is  i  =  45^o

Generally from Snell's law the angle of refraction of the blue light  in the acrylic block  is mathematically represented as

       r_F =  sin ^{-1}[\frac{sin(i) *  n_a }{n_F} ]

Where  n_a is the refractive index of air which have a value ofn_a =  1

So

     r_F =  sin ^{-1}[\frac{sin(45) *  1 }{ 1.497} ]

      r_F  =  28.18^o

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

       r_C =  sin ^{-1}[\frac{sin(i) *  n_a }{n_C} ]

Where  n_a is the refractive index of air which have a value ofn_a =  1

So

     r_C =  sin ^{-1}[\frac{sin(45) *  1 }{ 1.488} ]

      r_F  =  28.37^o

The angle between the blue beam and the red beam in the acrylic block

     \theta _d  =  r_C  - r_F

substituting values

       \theta _d  = 28.37 -  28.18

       \theta _d  =0.19 ^o

 

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