A thin film of water sits on top of a flat piece of glass, and light is incident from below (see the following figure). If the angle of incidence is 𝜃 = 28°, what is the angle of refraction of the final outgoing ray?
Question
A thin film of water sits on top of a flat piece of glass, and light is incident from below (see the following figure). If the angle of incidence is 𝜃 = 28°, what is the angle of refraction of the final outgoing ray?
Solution
To solve this problem, we need to use Snell's law, which states that the ratio of the sine of the angles of incidence and refraction is equivalent to the ratio of phase velocities (v1 / v2) in the two media, or the inverse of the ratio of the indices of refraction (n2 / n1).
n1 * sin(θ1) = n2 * sin(θ2)
Given that the light is incident from below, it first travels from glass to water. The refractive index of glass is typically around 1.5 and that of water is 1.33. The angle of incidence θ1 is given as 28°.
So, we can set up the equation as follows:
1.5 * sin(28°) = 1.33 * sin(θ2)
Solving for θ2 gives us the angle of refraction as the light exits the water and enters the air.
However, the problem asks for the final outgoing ray, which means we need to consider the refraction as the light exits the water and enters the air. The refractive index of air is approximately 1.
So, we use Snell's law again:
1.33 * sin(θ2) = 1 * sin(θ3)
Solving for θ3 will give us the angle of the final outgoing ray.
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