Question 1163329: If you are in shallow water and look at an object below the surface of the water,
the object will look farther away from you than it really is. This is because as light passes
through air and water it bends or refracts the light rays. The index of refraction for water is 1.33.
This is the ratio of the sine of theta 1 and the sine of theta 2.
a)You are in 2 feet of water in the shallow end of a pool. You look down at some goggles at angle
(theta=60) degrees measured from a line perpendicular to the surface of the water. Find (theta 2 )
b) Find the distances x and y
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(2264)   (Show Source):
You can put this solution on YOUR website! ```python?code_reference&code_event_index=2
import math
# Given values
n_air = 1.0
n_water = 1.33
theta_1_deg = 60.0
depth_y = 2.0 # feet
# Part a: Find theta_2
# n1 * sin(theta1) = n2 * sin(theta2)
theta_1_rad = math.radians(theta_1_deg)
sin_theta_2 = (n_air * math.sin(theta_1_rad)) / n_water
theta_2_rad = math.asin(sin_theta_2)
theta_2_deg = math.degrees(theta_2_rad)
# Part b: Find distances x and y
# y is the actual depth
y = depth_y
# x is the horizontal distance from the entry point to the object
# From the right triangle in the water: tan(theta_2) = x / y
x = y * math.tan(theta_2_rad)
print(f"{theta_1_deg=}")
print(f"{theta_2_deg=}")
print(f"{y=}")
print(f"{x=}")
```
```text?code_stdout&code_event_index=2
theta_1_deg=60.0
theta_2_deg=40.62813065148206
y=2.0
x=1.715911347328087
```
To solve for the angle of refraction and the geometric distances, we use **Snell's Law**, which describes how light bends when crossing the boundary between two different media (in this case, air and water).
### (a) Find $\theta_2$
Snell's Law is given by the formula:
$$n_1 \sin \theta_1 = n_2 \sin \theta_2$$
Where:
* $n_1$ is the index of refraction for air ($\approx 1.00$).
* $n_2$ is the index of refraction for water ($1.33$).
* $\theta_1$ is the angle of incidence in air ($60^\circ$).
* $\theta_2$ is the angle of refraction in water.
Using the ratio provided in the problem ($1.33 = \frac{\sin \theta_1}{\sin \theta_2}$):
$$1.33 = \frac{\sin 60^\circ}{\sin \theta_2}$$
$$\sin \theta_2 = \frac{\sin 60^\circ}{1.33}$$
$$\sin \theta_2 = \frac{0.8660}{1.33} \approx 0.6511$$
$$\theta_2 = \arcsin(0.6511) \approx 40.63^\circ$$
**The angle of refraction $\theta_2$ is approximately $40.63^\circ$.**
---
### (b) Find the distances $x$ and $y$
In this geometric model of refraction:
* **$y$** represents the vertical depth of the water. According to the problem, you are in 2 feet of water.
$$y = 2\text{ ft}$$
* **$x$** represents the horizontal distance from the point where the light ray enters the water to the location of the goggles on the pool floor.
Using the right triangle formed by the light ray under the water:
$$\tan \theta_2 = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{y}$$
$$x = y \cdot \tan \theta_2$$
$$x = 2 \cdot \tan(40.63^\circ)$$
$$x \approx 2 \cdot 0.8580$$
$$x \approx 1.716\text{ ft}$$
**Final Results:**
* **$\theta_2 \approx 40.63^\circ$**
* **$y = 2\text{ ft}$** (Actual depth)
* **$x \approx 1.72\text{ ft}$** (Horizontal distance)
Answer by ikleyn(53875)  (Show Source):
You can put this solution on YOUR website! .
If you are in shallow water and look at an object below the surface of the water,
the object will look farther away from you than it really is. This is because as light passes
through air and water it bends or refracts the light rays. The index of refraction for water is 1.33.
This is the ratio of the sine of theta 1 and the sine of theta 2.
a) You are in 2 feet of water in the shallow end of a pool. You look down at some goggles at angle
(theta=60) degrees measured from a line perpendicular to the surface of the water. Find (theta 2 )
b) Find the distances x and y
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
As the incoming post is worded, it is mathematically incorrect,
because is asks to find distances 'x' and 'y',
but does not determine what these distances are.
It is why this posted problem recalls me a lame horse with three legs.
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