A person looking into an empty container is able w sec the...

A person looking into an empty container is able w sec the far edge of the container's bottom, as shown in Figure P22.23a. The height of the container is $h$, and its width is $d$. When the container is completely filled with a fluid of index of refraction $n$, the person can see a coin at the middle of the container's bottom, as shown in Figure P22.23b. (a) Show that the ratio $h / d$ is given by $$ \frac{h}{d}=\sqrt{\frac{n^{2}-1}{4-n^{2}}} $$ (b) Assuming the container has a width of $8.0 \mathrm{~cm}$ and is filled with water, use the expression above to find the height of the container.

A person looking into an empty container is able w sec the far edge of the container's bottom, as shown in Figure P22.23a. The height of the container is $h$, and its width is $d$. When the container is completely filled with a fluid of index of refraction $n$, the person can see a coin at the middle of the container's bottom, as shown in Figure P22.23b. (a) Show that the ratio $h / d$ is given by
$$
\frac{h}{d}=\sqrt{\frac{n^{2}-1}{4-n^{2}}}
$$
(b) Assuming the container has a width of $8.0 \mathrm{~cm}$ and is filled with water, use the expression above to find the height of the container.

Key Concepts

Snell's Law

This principle relates the angles of incidence and refraction when a ray of light passes between two media of different refractive indices. It is defined by the equation n? sin(??) = n? sin(??), where n? and n? are the indices of refraction of the two media and ?? and ?? are the corresponding angles of incidence and refraction. It is used to determine the path that light takes when transitioning through an interface, which is fundamental in analyzing and solving problems involving refraction in fluids and glasses.

Critical Angle and Total Internal Reflection

The critical angle is the angle of incidence in the denser medium above which no refraction occurs, and instead all the light is reflected back internally. It is calculated using sin(?_c) = n?/n? when light passes from a medium of higher refractive index to one of lower refractive index. This concept is critical in optical systems where the limits of visibility or the confinement of light within a medium are determined by total internal reflection.

Geometrical Ray Analysis

This technique involves drawing ray diagrams and employing geometric relationships such as similar triangles to trace the paths of light rays. In optical problems, it allows one to relate different physical dimensions of the system, like heights and widths, based on the angular constraints imposed by refraction. This method is essential for deriving relationships between the geometry of optical containers and the behavior of light as it passes through different media.

Index of Refraction

The index of refraction is a dimensionless number that quantifies how much the speed of light is reduced inside a medium compared to its speed in a vacuum. It affects the bending of light at interfaces and is a key parameter in all equations involving Snell's Law and the determination of the critical angle, thereby influencing the overall design and analysis of optical systems.

Transcript

0:00 Hi there.

00:01 So for this problem, we are told that a person is looking into an empty container and is able to see the far edge of the container's bottom.

00:11 So the height of the container is age and its width is d.

00:18 This is its height and d is its width.

00:32 When the container is completely filled with a fluid of index of refruption end, the person can see a coin at the middle of the container's bottom.

00:43 So we need to show that the ratio h over d is equal to the square root of the interest of refraption n to the square minus 1, divided by 4 minus the index of refraption to the square.

01:07 So the situation that we have in here is the following.

01:16 This one right here, where a is when the container is empty, and b is when the container is filled.

01:26 So to solve this, we know that before the container is filled, the rates pass, as it's shown in this figure a at the left from this, and we can observe that the sign of theta that we call teta 1 is equal to this distance d over x1, which is the hypotenuse in this case.

01:59 So we will have d over x1.

02:05 And this is equal, we can write this in terms of the height h.

02:11 So it's going to be the definition of the hypotenuse that we know is the square root of h to the square plus d to the square.

02:24 So substituting in here what we are going to have for d, remember that d in this case, we're going to set that d is just, simply equal to, well, we can take d out of here, so we're going to have 1 over the square root of h over d plus 1, the square root of that.

02:57 So after the container is filled, the rates path that is shown in here in part b, we know that the distance d reduces to half of that.

03:09 So we're going to have a similar, a similar relationship, but in this case, we're going to have that sign of teta 2 is equal to d over 2 divided by x2, where we can write s2 as the square root of h to the square plus d over 2 to the square.

03:38 We take the square root of that.

03:41 Simplifying this expression, we will find that this is 1 over the square root of 4 times h over d to the square plus 1.

03:50 We take the square root of all of that.

03:53 So that's what we found...

Please give Ace some feedback