A laser beam is incident on a plate of glass that is 2.8 cm thick. The glass has an index of refraction of 1.6 and the angle of incidence is 36°. The top and bottom surfaces of the glass are parallel. What is the distance b between the beam formed by reflection off the top surface of the glass and the beam reflected off the bottom surface of the glass?

 b = 2.22 cm

Explanation:

The laser hits a point where the origin of the coordinate system is to carry out the measurements. When the ray enters the glass the angle of refraction is given by the equation

       n₁ sin θ₁ = n₂ sin θ₂

where n₁ is the index of refraction of air n₁ = 1 and n₂ is the index of refraction of glass n₂ = 1.6

       sin θ₂ = n₁ /n₂ sin θ₁

       sin θ₂ = 1 / 1.6 sin 36

       sin θ₂ = 0.367

        θ₂ = sin⁻¹ 0.367

        θ₂ = 21.6º

 with this angle and trigonometry we can find the distance x that the ray advances before reaching the bottom of the glass plate

        tan 21.6 = x / d

where d is the thickness of the glass d = 2.8 cm

         x = d tan 21.6

         x = 2.8 tan 21.6

         x = 1.11 cm

 as in the second surface it has a process of reflection the angle of reflection is equal to the angle of incidence θ_reflected = 21.6º, therefore to return to the upper surface recreate the same distance, therefore the total distance is

           b = 2x

           b = 2 1.11

           b = 2.22 cm

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