Concepts addressed: refraction, geometric and trigonometric analysis
Grade level: 10th.
A week ago, while walking through the park, I stood by the pond as I always do. While looking down at the water, I noticed something shiny – a golden coin! From where I stood, it seemed to be 1.50 m meters from the shore. I know I can only reach 1.00 m horizontally into the pond without getting my feet wet, but I also know that due to the refraction of light, the coin is not where it appears to be. As you can see in the image, I was standing on the edge of the pond, and my eyes are 1.70 m above the level of the water. The lake has a constant depth of 0.50 m deep.
- Is the coin closer or further from me than it seems?
- Can I reach the coin without getting my feet wet?
Refraction index air: n2 = 1
Refraction index water: n1 = 1.33
- Right triangle calculator – https://www.omnicalculator.com/math/right-triangle-side-angle
- Snell’s Law calculator – https://www.omnicalculator.com/physics/snells-law
- Complementary angles calculator – https://www.omnicalculator.com/math/complementary-angles
Question 1 hints:
Question 2 hints:
Step-by-Step solutions :
n1 * sin(θ1) = n2 * sin(θ2) => 1 * sin(θ1) = 1.33 * sin(θ2)
For the equality to hold, sin(θ1) > sin(θ2), and since both angles are smaller than 90º, θ1 > θ2 has to be true.
Look at the drawing. It means that the coin will be closer than it appears to be when I am looking from outside the water.
The first one is a triangle with hypotenuse going from my eyes to the image of the coin in the water. The other sides are known (my height plus the depth of the pond and the horizontal distance to the coin).
Using the Right triangle calculator, we can see that the angle I am looking at the coin with respect to the horizontal is: 56º
This is the same angle between my line of sight and the surface of the water. θ1 is, therefore, 90º – 56º = 34º
So, using the Snell’s Law calculator, we find θ2: 1 * sin (34º) = 1.33 * sin (θ1) => θ2 = 25º
This leaves us with another triangle, where the hypotenuse goes from the point my line of sight intersects water to the coin, with height being the depth of the water. We know one angle is 25º and the other is
90º – 25º = 65º
This will let us find the horizontal distance from the normal to the coin, which is 23 cm. We know the horizontal distance from me to the normal from the first triangle we constructed; it is 1.2 m
Adding both horizontal distances, we find the distance to the coin: 1.2 m + 0.23 m = 1.4 m. So I will not able to reach the coin.