A person swimming 0.80 m below the surface of the water in a...

Name: Problem 19, Chapter 34: University Physics with Modern Physics
Uploaded: 2019-08-05T19:58:21-08:00
Duration: 4 min 24 s
Channel: Bruce Edelman
Description: A person swimming 0.80 m below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is a height of 5.20 m above the swimmer. What is the actual height of the diving board above the surface of the water?

A person swimming 0.80 m below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is a height of 5.20 m above the swimmer. What is the actual height of the diving board above the surface of the water?

Key Concepts

Snell's Law

Snell's Law describes the relationship between the angles of incidence and refraction and the refractive indices of the two media involved. It provides the mathematical basis for understanding how light bends at an interface, which in turn helps in determining the transformation between the actual and apparent locations of objects.

Apparent Depth

Apparent depth is the perceived position of an object when viewed across a refractive boundary. Due to refraction, objects located in one medium may seem to be at a different depth than they actually are. This concept is crucial in calculating the real distance or height of an object when its image position has been altered by the refractive interface.

Refraction

Refraction is the bending of light as it passes from one medium into another with a different refractive index. This change in direction occurs because light travels at different speeds in different media, leading to a shift in the apparent position of objects when viewed across the interface.

Refractive Index

The refractive index is a measure of how much a medium slows down light compared to its speed in a vacuum. It determines the degree to which light will bend when it passes between substances, and in problems involving water and air, the indices (typically about 1.33 for water and 1.00 for air) are used to relate the actual position of objects to their apparent positions.

Transcript

00:01 Okay, so we're doing chapter 34 problem 19 here.

00:07 And so this problem says a person swimming 0 .8 meters below the surface.

00:12 And i have this drawing here, so let's draw this as 0 .8 meters below the surface of the water.

00:20 And it says it's, you're looking at the diving board that is directly overhead, and the person sees an image of the board that is formed by refraction at the surface of the water.

00:32 This image is a height of 5 .2 meters above the swimmer and what is the actual height of the diving board? so i just drew this as to scale or just to show that there is an image formed on the other side of the surface.

00:52 So we know the image is negative.

00:58 But i drew it as the image being closer, but we don't actually know that.

01:03 So we're going to have to use our equation to figure out.

01:05 We could use some geometry, but we can also just use our usual sign convention and our lens equations to figure this out.

01:12 So don't take that as meaning the image is necessarily going to be closer or not.

01:16 We have not figured that out yet.

01:20 But let's go ahead and label where these distances.

01:24 So this would be our object distance.

01:26 And this over here, s prime, will be our image distance.

01:30 So it tells us that s prime is 5 .2 meters or negative 5 .2 meters since it's on the same side as the object.

01:40 Now when we're talking about the surface of the water here, we can think that as a plain lens, so it has an infinity radius of curvature, which is going to be important.

01:52 So now we can plug this into our equation for a lens on two different.

02:02 Between two different mediums.

02:04 And what we're left with here is we see that the in of the...

02:15 Oh yeah, we should also mention that this top interface is the air, right? n -a, and this is water down here.

02:22 And n -a is one and water is four -thirds.

02:26 So our equation then becomes the a over the object distance...

Please give Ace some feedback