∙A zoo aquarium has transparent walls, so that spectators on...

$\bullet \mathrm{A}$ zoo aquarium has transparent walls, so that spectators on both sides of it can watch the fish. The aquarium is 5.50 $\mathrm{m}$ across, and the spectators on both sides of it are standing 1.20 $\mathrm{m}$ from the wall. How far away do spectators on one side of the aquarium appear to those on the other side? (Ignore any refraction in the walls of the aquarium.)

$\bullet \mathrm{A}$ zoo aquarium has transparent walls, so that spectators on both sides of it can watch the fish. The aquarium is 5.50 $\mathrm{m}$ across, and the spectators on both sides of it are standing 1.20 $\mathrm{m}$ from the wall. How far away do spectators on one side of the
aquarium appear to those on the other side? (Ignore any refraction in the walls of the aquarium.)

Key Concepts

Straight?Line Propagation of Light

In geometrical optics, light is assumed to travel in straight lines through a homogeneous medium when no refraction or reflection is involved. This assumption lets us use simple geometric constructions to analyze how light travels from an object to an observer, which is essential for understanding apparent positions when viewing objects through transparent interfaces.

Virtual Image Formation at Plane Interfaces

When light passes through a transparent, flat interface without being refracted, the observer perceives objects as if they are located along the straight line defined by the last interface encountered. The concept of a virtual image—that is, an image located at a distance behind the interface equal to the object’s distance in front—is key to predicting the apparent location of objects viewed through such surfaces.

Application of Similar Triangles in Optical Geometry

Similar triangles provide a useful method for relating the distances in an optical system. By constructing triangles between the observer, the interface, and the object, one can derive relationships that determine the apparent distance or displacement of an object. This method is particularly useful in problems where the geometry of the setup (such as distances from the walls or interfaces) influences how the object’s image is perceived.

Transcript

00:01 Okay, so for this problem, we want to consider that we have this aquarium and it has a certain length, and then there are viewers in both sides, and then we want to figure out what the perceived distances between the viewers.

00:15 And so basically what we want to do is we want to figure out how a sub question we want to answer is how long the tank appears to either viewer, and then we can add that to the distance between the viewers.

00:27 And pause this.

00:31 I'm having some trouble with my tablet.

00:33 Okay, great.

00:34 My tablet's working.

00:37 So let's answer the question of finding the apparent depth by considering, let's say, you could think of it as considering an object, see here, and you're here, and you want to figure out where the object appears relative.

01:01 Where the image of this object at the other end would appear to get the effective length of the tank.

01:10 So the, oops.

01:16 So the image position is at 5 .5 meters.

01:24 So s is equal to 5 .5.

01:27 And the index of refraction where the object is is 1 .33 because it's in water.

01:33 Put the dot down here.

01:41 And we want to find s prime where the index of refraction is air.

01:47 So that's nb equals 1.

01:50 And then the equation we would use to do this is the equation that most generally describes having a curve surface leading an index of refract, an object with some difference in index of refraction.

02:06 I mean, yeah, i wish there was a nice name for this equation, but it's 24 .11, or if there is one, i wish i knew it.

02:15 But yeah, i'm just pulling up the book now so i can copy it.

02:20 I don't have it memorized...

Please give Ace some feedback