Problem 46
Question
You are submerged in a swimming pool. What is the maximum angle at which you can see light coming from above the pool surface? That is, what is the angle for total internal reflection from water into air?
Step-by-Step Solution
Verified Answer
Answer: The maximum angle at which light can be seen from above the surface of a swimming pool is approximately 48.8°, which is the angle for total internal reflection from water to air.
1Step 1: Find the refractive indices of the mediums
Given that we are considering light traveling from water (medium 1) to air (medium 2), we need to find the refractive indices of these two mediums. The refractive index of water is approximately 1.33 and that of air is 1.00.
2Step 2: Understand the Snell's Law formula
Snell's law relates the angle of incidence, angle of refraction, and the refractive indices of the mediums involved. It is given by the formula:
n_1 * sin(theta_1) = n_2 * sin(theta_2)
Here, n_1 and n_2 are the refractive indices of medium 1 and medium 2, respectively, while theta_1 is the angle of incidence and theta_2 is the angle of refraction.
3Step 3: Set up the formula for critical angle
At the critical angle, the angle of refraction (theta_2) is 90 degrees. Plugging this into Snell's law and rearranging for the angle of incidence (theta_1), we get:
sin(theta_1) = n_2 * sin(90°) / n_1
4Step 4: Apply values and calculate critical angle
Substitute the refractive indices of water and air found in Step 1:
sin(theta_1) = 1.00 * sin(90°) / 1.33
The sine of 90 degrees is 1, so the equation simplifies to:
sin(theta_1) = 1 / 1.33
Now, find the inverse sine to get the critical angle:
theta_1 = arcsin(1 / 1.33) ≈ 48.8°
5Step 5: Interpret the results
The maximum angle at which you can see light coming from above the pool surface is equal to the critical angle, which we calculated to be approximately 48.8°. This is the angle for total internal reflection from water to air in a swimming pool, and represents the threshold at which light entering from above the surface can no longer be seen by an observer submerged in the water.
Other exercises in this chapter
Problem 44
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