Answer:
The angle between the blue beam and the red beam in the acrylic block is
[tex]\theta _d =0.19 ^o[/tex]
Explanation:
From the question we are told that
The refractive index of the transparent acrylic plastic for blue light is [tex]n_F = 1.497[/tex]
The wavelength of the blue light is [tex]F = 486.1 nm = 486.1 *10^{-9} \ m[/tex]
The refractive index of the transparent acrylic plastic for red light is [tex]n_C = 1.488[/tex]
The wavelength of the red light is [tex]C = 656.3 nm = 656.3 *10^{-9} \ m[/tex]
The incidence angle is [tex]i = 45^o[/tex]
Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as
[tex]r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ][/tex]
Where [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]
So
[tex]r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ][/tex]
[tex]r_F = 28.18^o[/tex]
Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as
[tex]r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ][/tex]
Where [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]
So
[tex]r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ][/tex]
[tex]r_F = 28.37^o[/tex]
The angle between the blue beam and the red beam in the acrylic block
[tex]\theta _d = r_C - r_F[/tex]
substituting values
[tex]\theta _d = 28.37 - 28.18[/tex]
[tex]\theta _d =0.19 ^o[/tex]
