The average index of refraction of water is n=1.336.
(a) What is the Brewster angle θBr when light reflects in water off air (i.e. at point B) (give your answer in degrees)?
(b) r is the angle of incidence of the light ray at point B in the figure above. What is r−θBr (give your answer in degrees)?
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To solve these questions, we need to use Snell's law and the law of reflection. Snell's law relates the angles of incidence and refraction when light passes through a boundary between two different media, while the law of reflection describes the angle of reflection when light bounces off a surface.
(a) To determine the Brewster angle θBr, we can use Snell's law. The Brewster angle is the angle of incidence at which light undergoes pure polarization (reflection or transmission) when it reflects off a surface. In this case, the light passes from water to air.
Snell's law states: n₁sinθ₁ = n₂sinθ₂,
Where n₁ and n₂ are the indices of refraction for the two media, and θ₁ and θ₂ are the angles of incidence and refraction, respectively.
Given:
n₁ = 1.336 (index of refraction of water)
n₂ = 1 (index of refraction of air)
We need to find θBr when light reflects in water off air (i.e., at point B).
Since θ₂ is the angle of refraction in air and light undergoes total internal reflection at the Brewster angle, we can substitute n₂ = 1 and sinθ₂ = 1 (from air to air) into Snell's law:
n₁sinθ₁ = n₂sinθ₂,
1.336sinθBr = 1(1),
1.336sinθBr = 1.
Now, we solve for θBr:
sinθBr = 1/1.336,
θBr = arcsin(1/1.336),
Using a calculator to find the inverse sine (arcsin) of 1/1.336, we get θBr ≈ 48.7 degrees.
So, the Brewster angle θBr is approximately 48.7 degrees.
(b) Now we need to calculate r - θBr, where r is the angle of incidence of the light ray at point B in the figure. Given the diagram, r can be identified as the angle between the normal line and the incident ray of light. The Brewster angle θBr, obtained in part (a), is a fixed value.
We can use the law of reflection, which states that the angle of incidence is equal to the angle of reflection.
Therefore, r - θBr is the angle between the reflected ray (line BA) and the normal line.
Since the angle of reflection (r) is equal to the angle of incidence (r - θBr), we have:
r - θBr = r.
So, r - θBr is equal to r.
Therefore, r - θBr is also approximately 48.7 degrees.