A tank whose bottom is a mirror is filled with water to a...

Name: Problem 16, Chapter 34: University Physics with Modern Physics
Uploaded: 2019-04-16T18:27:24-08:00
Duration: 6 min 45 s
Channel: Ryan Kutayiah
Description: A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence?

A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence?

Key Concepts

Normal Incidence

Normal incidence refers to the situation where light rays hit a surface perpendicularly, or at an angle very close to 90 degrees with respect to the surface. In such cases, the effects of refraction are minimized, simplifying the calculation of apparent depth and other optical phenomena.

Refraction

When light travels from one medium to another with a different refractive index, its speed and direction change. This bending of light is described by Snell's Law and is responsible for the apparent positional shift of objects submerged in a medium like water. It is fundamental to understanding why objects appear shallower than they actually are.

Apparent Depth

Apparent depth is the perceived depth of an object under water when viewed from above. Due to the refraction of light at the water-air interface, the light rays bend, making the object look closer to the surface than its true physical position. This concept is key in analyzing observations in optical scenarios.

Reflection from a Mirror

Reflection from a mirror involves light bouncing off a surface such that the angle of incidence equals the angle of reflection. When a mirror is present, it creates a virtual image that appears behind the reflective surface at an equal distance, influencing the observer's perception of the object's position.

Transcript

00:01 In problem 16, we're told we have a tank with a mirror at the bottom, and the tank is filled with water to a depth of 20 centimeters.

00:11 So this blue line is the water line, the water air interface.

00:15 And we have a small fish that is floating motionless, seven centimeters below the water's surface.

00:23 So this green line here is my fish, and it's seven centimeters below the water.

00:30 And in part a of the problem, we're asked to find the apparent depth of the fish when viewed at normal incidents.

00:37 So at normal incidents, you're looking straight down here.

00:42 So basically what's happening, there's light coming from the fish to your eyes, which enables you to see it, but the fish is in water and your eyes are in air.

00:53 And so that will make the fish appear at a different distance than its actual distance because the light is traveling through two different media with different indices of a fraction.

01:07 So the way we'll solve this problem is to use equation 34 .13.

01:16 And this is the equation for a plain mirror when you are in two different medium.

01:26 So we have n .a, which is a fraction of medium a over s plus nb over s primed.

01:42 And actually i mislabeled my a and b in the figures here.

01:47 So actually, that should be b and this is a.

01:54 Okay.

01:55 So n a is water and nb is air.

02:01 And that's equal to zero.

02:05 And so we know n -a, we know nb, we know s.

02:10 So s is the object distance, and s -prime is the apparent distance.

02:18 So we know that the object distance is seven centimeters...