Building contractors often install double-glazed windows to prevent thermal energy (heat) from entering or exiting a building. In addition to being effective insulators, such windows present interesting optical effects. In the figure, a double-glazed window consists of two identical panes of glass ( g=1.50 ), each g=42.0 mm thick, separated by an air gap of a=33.6 mm . If light incident on the glass makes an angle of =40.00∘ with respect to the glass, find the shift in path Δ as the light enters the room. Use a=1.00 for the index of refraction of air.

Two parallel, horizontal rectangular regions representing glass panes, separated by an open rectangular region representing the air gap between the glass. The thickness of each glass pane is y subscript g. The thickness of the air gap is y subscript a. A light ray is incident upon the upper glass pane, and makes an angle phi with respect to the glass surface. The ray bends downward toward the normal at the air-glass boundary. The ray passes through the glass and strikes the bottom glass-air boundary, where it bends away from the normal and enters the air gap. The beam continues to the second glass sheet, where it once again bends toward the normal at the air-glass boundary. The ray passes through the second sheet and again bends away from the normal at the glass-air boundary. The ray that emerges from the second sheet propagates parallel to a straight line extending the trajectory of the ray incident on the first glass sheet, and is a perpendicular distance delta x away from it.
Δ=
mm

  x_total = 0.46 cm

Explanation:

Let's analyze this exercise we can calculate the lateral displacement formed by the displacement of the glass plate plus the displacement in the part of air between the glasses.

       x_total = d₁ - d₂ + d₃

The sign of displacement in the air is because it reduces lateral displacement.

Let's use the law of refraction to find the angle refracted in the glass, which will determine the displacement

          n₁ sin θ₁ = n₂ sin θ₂

where subscript 1 is for the indicating part and subscript 2 for the refracted part (glass)

         sin θ₂ = n₁ /n₂  sin θ₁

         sin θ₂ = 1 /n_g sin θ₁

for the specific case

         sin θ₂ = 1 /1.5 sin  40

         θ₂ = sin⁻¹ (0.4285)

         θ₂ = 25.37º

so the lateral displacement when exiting the plate is

        tan θ₂ = d₁ / y_g

        d₁ = y_g tan 25.37

        d₁ = 4.20 so 25.37

        d₁ = 1.99 cm

when the ray leaves glass and enters the air, it will recover its initial direction, therefore the angle is   θ = 40º, the displacement for this part is

     tan θ = d₂ / y_a

     d₂ = y_a sin θ

suppose that the width is equal to the thickness of the glass  y_a = 4.20 cm

     d₂ = 4.20 tan 40

     d₂ = 3.52 cm

At this time it enters the second glass plate, we see that it is identical to the analysis of the first plate. So the displacement is

        d₃ = y_g tan 25.37

        d₃ = 4.2 tan 2537

        d₃ = 1.99 cm

consequently the total displacement is

general fromula

       x_total = 2 y_g tan 25.37 - y_a tan 40

The offset for the given values is

      x_total = 1.99 - 3.52 + 1.99

      x_total = 0.46 cm

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.