When a ray of light traveling in air hits a tilted plane parallel slab (of glass, say), it emerges parallel to the original ray but shifted transversely. Carefully draw out the situation and use Snell’s law to derive the amount of the transverse shift, x, as a function of the tilt angle of the slab, θ, its thickness, d, and its index of refraction, n. Find the exact expression with no approximations. We recommend you do this out all in variables because it's a useful formula to have. Also, you will want this for the following questions. However, since the auto-grader has difficulty with these formulas, use n=1.5, d=1.0 cm, and θ = 45° and enter a numerical answer. Give your answer in cm to two significant figures.

  x =  0.4654 cm

Explanation:

In this exercise we use the law of refraction

           n₁ sin θ₁ = n₂ sin θ₂

apply this formula to the first surface, where n₁ is the index of refraction of air (n₁ = 1) and n₂ is the index of refraction of glass (n₂ = n)

            θ₂ = sin⁻¹ (sin θ₁ / n)         (1)

having this angle we use trigonometry to find the value of the point where it comes out when we reach the other side

refracted ray

            tan θ₂ = x₂ / d

            x₂ = d tan θ₂

this value is the distance displaced by the refracted ray

now let's find the distance at which the incident beam should exit

           tan θ₁ = x₁ / d

           x₁ = d tan θ₁

the displacement of the ray is the difference between these two distances, we will call it x

           x = x₁ - x₂

            x = d tan θ₁ - d tan θ₂

           x = d (tan θ₁ - tan θ₂)        (2)

the easiest way to do the calculations is to find tea2 from the binding 1 and then perform the calculation with equation 2

calculate

            θ₂ = sin⁻¹ (sin 45 /1.5)

             θ₂ = 28.13º

             x = 1.0 (tan 45 - tan 28.13)

             x =  0.4654 cm

50 inches should be the resultant vector