A large fish tank is filled with water (of refractive index 1.33). One side of the tank is made of a parallel-sided piece of glass. A ray of light strikes the glass from the outside at an angle of incidence of 48.
Estimate the angle through which this ray is deviated as it emerges into the water. Just give the magnitude of the deviation, there's no need for signs.
Give your answer in degrees.
Estimate? 10 degrees.
Why not calculate it?
I don't know how to calculate it, that's why I was looking for help
Because the glass side wall is plane-parallel, Snell's law takes the form
N1*sinA1 = N2*sinA2 = N3*sinA3, where
N1 = index in air = 1.00: A1 is angle of incidence in air = 48 degrees
N2 = refractive index in glass = ?
N3 = refractive index in water = 1.33; A3 = angle of emergence (relative to normal) in water.
You can ignore what happens in the glass and write:
1.33*sinA3 = 1.00*sinA1
SinA3 = 0.559
A3 = 34.0 degrees
Angle of deviation (air to water)
= A1 - A3 = 14.0 degrees
To estimate the angle through which the light ray is deviated as it emerges into the water, we can use Snell's Law which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the refractive indices of the two media.
Let's denote the angle of incidence of the light ray on the glass as θ1 and the angle of refraction within the glass as θ2. The refractive index of the glass is not given, so we'll assume it is the standard value of 1.50.
Since the light ray is coming from outside (air) to inside the glass, Snell's Law can be written as:
sin(θ1) / sin(θ2) = n2 / n1.
We need to find the angle of refraction within the glass. Rearranging the equation, we get:
sin(θ2) = (n1 / n2) * sin(θ1).
Substituting the values into the equation and solving for θ2:
θ2 = arcsin((n1 / n2) * sin(θ1)).
Given that the angle of incidence (θ1) is 48 degrees, the refractive index of water (n1) is 1.33, and the refractive index of glass (n2) is 1.50, we can calculate the angle of refraction (θ2).
θ2 = arcsin((1.33 / 1.50) * sin(48)).
Using a calculator, we find that θ2 is approximately 39.95 degrees.
Therefore, the magnitude of the angle by which the light ray is deviated as it emerges into the water is approximately 39.95 degrees.