# Browsing Animations: Optics, Geometric

### apparent-depth-6.iwp

An alien on shore observes a neutrally-buoyant object (orange) in the water. The angle subtended by the refracted rays at the alien's eye is shown in yellow. The apparent position of the object is shown in gray. The refracted rays from the alien's eye are extended backward into the water. Playing the animation will move the object the right. Unphysical behavior may be shown if the object moves too far to the right. The magnification is the ratio of the angle subtended at the eye by the refracted rays and the angle that would be subtended in the absence of water. A graph of magnification as a function of time can be displayed by clicking on Show graph. Each unit of time represents a horizontal displacement of 0.1 grid unit. Note that magnification is not the ratio of image to object size. It is, however, the ratio of the component of the image diameter perpendicular to the line of sight to the corresponding component of the object diameter.

### apparent-depth-template.iwp

An observer at upper right views a neutrally-buoyant object (orange) in the water. The angle subtended by the refracted rays at the observer's eye is shown in yellow. The apparent position of the object is shown in gray. The refracted rays from the observer's eye are extended backward into the water. Playing the animation will move the object the right. Unphysical behavior may be shown if the object moves too far to the right. The magnification is the ratio of the angle subtended at the eye by the refracted rays and the angle that would be subtended in the absence of water. A graph of magnification as a function of time can be displayed by clicking on Show graph. Each unit of time represents a horizontal displacement of 0.1 grid unit. Note that magnification is not the ratio of image to object size. It is, however, the ratio of the component of the image diameter perpendicular to the line of sight to the corresponding component of the object diameter. Notes for instructor: Apparent rays are constructed on the assumption that the apparent distance to the right-hand side of the image is inversely proportional to the angle subtended at the eye by the refracted rays. (This takes the diameter of the object to be a constant. Basically, we're looking for the location at which the object in air would subtend an angle equal to the angle subtended by the refracted rays.) 1 refers to the rays (incident, refracted, apparent from the left side of the object). 2 refers to the rays (incident, refracted, apparent from the right side of the object). The X-intercept is the point where the ray incident from water refracts into air. These coordinates are determined using the principle of least time. The equation resulting from the application of the principle is a quartic, which is solved for the applicable real root. Notation: (xo,yo) = coordinates of observer (x1,y1) = coordinates of left side of object (x2,y1) = coordinates of right side of object (z,0) = coordinates of left ray at boundary (zd,0) = coordinates of right ray at boundary

### least-time-6.iwp

The lower half of the screen is water, and the upper half is air. Running the animation plots paths of rays from an object in the lower left-hand corner of the screen to an observer in the upper right-hand corner. Physically, only one of the infinite number of paths is possible. That one path is the one for which the ray takes the least time from object to observer. Click on Show graph to display the travel times in air (red), water (blue), and the total travel time (black) as a function of the x-axis coordinate where the path bends.* The value of x for which the total time is a minimum is the path that light actually takes. Since the minimum is fairly broad, it's difficult to determine the corresponding value of x accurately. One shouldn't assume that the minimum occurs where the times of travel in the air and the water are the same. A more accurate predictor is the violet line. The zero of this function is the value of x corresponding to the least total time of travel and also the value of x for which Snell's holds. The zero of the function mentioned above is also calculated analytically and is given as the output labeled Quartic root. *The time scale in the graph is related to the x-coordinate of the bend as follows: x = (initial bend coordinate) + (x-step)(time). Note to authors: The solution of the quartic differs slightly from the least-time method. This is an unresolved discrepancy. Using the quartic root to calculate the angles of incidence and refraction does yield values that obey Snell's Law. Perhaps there is a problem in the time calculation.

### apparent-depth-template.iwp

An observer at upper right views a neutrally-buoyant object (orange) in the water. The angle subtended by the refracted rays at the observer's eye is shown in yellow. The apparent position of the object is shown in gray. The refracted rays from the observer's eye are extended backward into the water. Playing the animation will move the object the right. Unphysical behavior may be shown if the object moves too far to the right. The magnification is the ratio of the angle subtended at the eye by the refracted rays and the angle that would be subtended in the absence of water. A graph of magnification as a function of time can be displayed by clicking on Show graph. Each unit of time represents a horizontal displacement of 0.1 grid unit. Note that magnification is not the ratio of image to object size. It is, however, the ratio of the component of the image diameter perpendicular to the line of sight to the corresponding component of the object diameter. Notes for instructor: Apparent rays are constructed on the assumption that the apparent distance to the right-hand side of the image is inversely proportional to the angle subtended at the eye by the refracted rays. (This takes the diameter of the object to be a constant. Basically, we're looking for the location at which the object in air would subtend an angle equal to the angle subtended by the refracted rays.) 1 refers to the rays (incident, refracted, apparent from the left side of the object). 2 refers to the rays (incident, refracted, apparent from the right side of the object). The X-intercept is the point where the ray incident from water refracts into air. These coordinates are determined using the principle of least time. The equation resulting from the application of the principle is a quartic, which is solved for the applicable real root. Notation: (xo,yo) = coordinates of observer (x1,y1) = coordinates of left side of object (x2,y1) = coordinates of right side of object (z,0) = coordinates of left ray at boundary (zd,0) = coordinates of right ray at boundary

### least-time-6.iwp

The lower half of the screen is water, and the upper half is air. Running the animation plots paths of rays from an object in the lower left-hand corner of the screen to an observer in the upper right-hand corner. Physically, only one of the infinite number of paths is possible. That one path is the one for which the ray takes the least time from object to observer. Click on Show graph to display the travel times in air (red), water (blue), and the total travel time (black) as a function of the x-axis coordinate where the path bends.* The value of x for which the total time is a minimum is the path that light actually takes. Since the minimum is fairly broad, it's difficult to determine the corresponding value of x accurately. One shouldn't assume that the minimum occurs where the times of travel in the air and the water are the same. A more accurate predictor is the violet line. The zero of this function is the value of x corresponding to the least total time of travel and also the value of x for which Snell's holds. The zero of the function mentioned above is also calculated analytically and is given as the output labeled Quartic root. *The time scale in the graph is related to the x-coordinate of the bend as follows: x = (initial bend coordinate) + (x-step)(time). Note to authors: The solution of the quartic differs slightly from the least-time method. This is an unresolved discrepancy. Using the quartic root to calculate the angles of incidence and refraction does yield values that obey Snell's Law. Perhaps there is a problem in the time calculation.

### prism-1.iwp

A ray of light is incident from air on a prism with index of refraction 1.5. Playing the animation will increase the angle of incidence in 1 degree increments. Normals to the sides of the prism are indicated by blue lines. The angle of incidence is given as an output. For an angle of incidence 15 degrees, use Snell's Law and the geometry of the prism to determine the angle of refraction at which the ray emerges from the prism on the right.

### least-time-6.iwp

The lower half of the screen is water, and the upper half is air. Running the animation plots paths of rays from an object in the lower left-hand corner of the screen to an observer in the upper right-hand corner. Physically, only one of the infinite number of paths is possible. That one path is the one for which the ray takes the least time from object to observer. Click on Show graph to display the travel times in air (red), water (blue), and the total travel time (black) as a function of the x-axis coordinate where the path bends.* The value of x for which the total time is a minimum is the path that light actually takes. Since the minimum is fairly broad, it's difficult to determine the corresponding value of x accurately. One shouldn't assume that the minimum occurs where the times of travel in the air and the water are the same. A more accurate predictor is the violet line. The zero of this function is the value of x corresponding to the least total time of travel and also the value of x for which Snell's holds. The zero of the function mentioned above is also calculated analytically and is given as the output labeled Quartic root. *The time scale in the graph is related to the x-coordinate of the bend as follows: x = (initial bend coordinate) + (x-step)(time). Note to authors: The solution of the quartic differs slightly from the least-time method. This is an unresolved discrepancy. Using the quartic root to calculate the angles of incidence and refraction does yield values that obey Snell's Law. Perhaps there is a problem in the time calculation.

### prism-1.iwp

A ray of light is incident from air on a prism with index of refraction 1.5. Playing the animation will increase the angle of incidence in 1 degree increments. Normals to the sides of the prism are indicated by blue lines. The angle of incidence is given as an output. For an angle of incidence 15 degrees, use Snell's Law and the geometry of the prism to determine the angle of refraction at which the ray emerges from the prism on the right.

### prism-3b.iwp

A ray of light is incident from air on a prism with the given index of refraction. Normals to the sides of the prism are shown in red. Playing the animation will increase the vertex angle of the prism in 1 degree increments. (If the critical angle is exceeded for the ray incident on the right side of the prism, non-physical results may occur.) Run the animation without changing anything. Then set the index of refraction to 1 and play the animation. Explain why the ray is undeviated independent of the vertex angle.

### prism-1.iwp

A ray of light is incident from air on a prism with index of refraction 1.5. Playing the animation will increase the angle of incidence in 1 degree increments. Normals to the sides of the prism are indicated by blue lines. The angle of incidence is given as an output. For an angle of incidence 15 degrees, use Snell's Law and the geometry of the prism to determine the angle of refraction at which the ray emerges from the prism on the right.

### prism-3b.iwp

A ray of light is incident from air on a prism with the given index of refraction. Normals to the sides of the prism are shown in red. Playing the animation will increase the vertex angle of the prism in 1 degree increments. (If the critical angle is exceeded for the ray incident on the right side of the prism, non-physical results may occur.) Run the animation without changing anything. Then set the index of refraction to 1 and play the animation. Explain why the ray is undeviated independent of the vertex angle.

### prism-6.iwp

The dispersion of white light by crown glass is modeled. The index of refraction ranges from 1.513 for red light to 1.532 for violet light. Playing the animation will decrease the angle of incidence in 1 degree increments. The angles of refraction for red and violet light coming out of the prism are given as outputs. Use Snell's Law and the geometry of the prism to verify that the angles are correct for an angle of incidence that you select.

### prism-3b.iwp

A ray of light is incident from air on a prism with the given index of refraction. Normals to the sides of the prism are shown in red. Playing the animation will increase the vertex angle of the prism in 1 degree increments. (If the critical angle is exceeded for the ray incident on the right side of the prism, non-physical results may occur.) Run the animation without changing anything. Then set the index of refraction to 1 and play the animation. Explain why the ray is undeviated independent of the vertex angle.

### prism-6.iwp

The dispersion of white light by crown glass is modeled. The index of refraction ranges from 1.513 for red light to 1.532 for violet light. Playing the animation will decrease the angle of incidence in 1 degree increments. The angles of refraction for red and violet light coming out of the prism are given as outputs. Use Snell's Law and the geometry of the prism to verify that the angles are correct for an angle of incidence that you select.

### ray-refraction-3b.iwp

The blue and gray areas represent media of different optical densities. n2/n1 is the ratio of the optical density of the gray medium to that of the blue medium. The normal to the boundary is shown in red. The path of a light ray is shown in yellow. The ray is incident from the blue medium. Playing the applet forward or backwards will increase or decrease the angle of incidence. Explain why the refracted ray disappears at a particular angle. Use Snell's Law to show that the angle at which the refracted ray disappears is correct. Values of the angles of incidence and refraction are given as outputs.

### prism-6.iwp

The dispersion of white light by crown glass is modeled. The index of refraction ranges from 1.513 for red light to 1.532 for violet light. Playing the animation will decrease the angle of incidence in 1 degree increments. The angles of refraction for red and violet light coming out of the prism are given as outputs. Use Snell's Law and the geometry of the prism to verify that the angles are correct for an angle of incidence that you select.

### ray-refraction-3b.iwp

The blue and gray areas represent media of different optical densities. n2/n1 is the ratio of the optical density of the gray medium to that of the blue medium. The normal to the boundary is shown in red. The path of a light ray is shown in yellow. The ray is incident from the blue medium. Playing the applet forward or backwards will increase or decrease the angle of incidence. Explain why the refracted ray disappears at a particular angle. Use Snell's Law to show that the angle at which the refracted ray disappears is correct. Values of the angles of incidence and refraction are given as outputs.

### ray-refraction-3c.iwp

The blue and gray areas represent media of different optical densities. n2/n1 is the ratio of the optical density of the gray medium to that of the blue medium. The normal to the boundary is shown in red. The path of a light ray is shown in yellow. The ray is incident from the blue medium. Playing the applet forward or backwards will increase or decrease the angle of incidence. Explain why the refracted ray disappears at a particular angle. Use Snell's Law to show that the angle at which the refracted ray disappears is correct. Values of the angles of incidence and refraction are given as outputs.

### ray-refraction-3b.iwp

The blue and gray areas represent media of different optical densities. n2/n1 is the ratio of the optical density of the gray medium to that of the blue medium. The normal to the boundary is shown in red. The path of a light ray is shown in yellow. The ray is incident from the blue medium. Playing the applet forward or backwards will increase or decrease the angle of incidence. Explain why the refracted ray disappears at a particular angle. Use Snell's Law to show that the angle at which the refracted ray disappears is correct. Values of the angles of incidence and refraction are given as outputs.

### ray-refraction-3c.iwp

The blue and gray areas represent media of different optical densities. n2/n1 is the ratio of the optical density of the gray medium to that of the blue medium. The normal to the boundary is shown in red. The path of a light ray is shown in yellow. The ray is incident from the blue medium. Playing the applet forward or backwards will increase or decrease the angle of incidence. Explain why the refracted ray disappears at a particular angle. Use Snell's Law to show that the angle at which the refracted ray disappears is correct. Values of the angles of incidence and refraction are given as outputs.

### ray-refraction-3d.iwp

The red line at y = 0 represents a boundary between media of different indices of refraction. The path of a light ray is shown in blue. Playing the applet forward or backwards will increase or decrease the angle of incidence. Determine the ratio of the index of refraction of the upper medium (above the boundary) to that of the lower medium (below the axis.)

### ray-refraction-3c.iwp

The blue and gray areas represent media of different optical densities. n2/n1 is the ratio of the optical density of the gray medium to that of the blue medium. The normal to the boundary is shown in red. The path of a light ray is shown in yellow. The ray is incident from the blue medium. Playing the applet forward or backwards will increase or decrease the angle of incidence. Explain why the refracted ray disappears at a particular angle. Use Snell's Law to show that the angle at which the refracted ray disappears is correct. Values of the angles of incidence and refraction are given as outputs.

### ray-refraction-3d.iwp

The red line at y = 0 represents a boundary between media of different indices of refraction. The path of a light ray is shown in blue. Playing the applet forward or backwards will increase or decrease the angle of incidence. Determine the ratio of the index of refraction of the upper medium (above the boundary) to that of the lower medium (below the axis.)

### ray-refraction-3e.iwp

The red line at y = 0 represents a boundary between different media. The path of a light ray is shown in blue. Playing the applet forward or backwards will change the angle of incidence. Which ray is the incident ray? Which is refracted? reflected? In which medium is the speed of light greater? Suppose wavefronts were added to the rays. For which two rays would the wavelength be the same? How would the wavelengths in the 2 media compare? Why?

### ray-refraction-3d.iwp

The red line at y = 0 represents a boundary between media of different indices of refraction. The path of a light ray is shown in blue. Playing the applet forward or backwards will increase or decrease the angle of incidence. Determine the ratio of the index of refraction of the upper medium (above the boundary) to that of the lower medium (below the axis.)

### ray-refraction-3e.iwp

The red line at y = 0 represents a boundary between different media. The path of a light ray is shown in blue. Playing the applet forward or backwards will change the angle of incidence. Which ray is the incident ray? Which is refracted? reflected? In which medium is the speed of light greater? Suppose wavefronts were added to the rays. For which two rays would the wavelength be the same? How would the wavelengths in the 2 media compare? Why?

### ray-refraction-3f.iwp

The blue and gray areas represent different media. Incident, reflected, and refracted light rays in the media are shown. In which medium is the speed of light greater? How do you know? Run the applet to increase the angle of incidence. What will happen when the angle of refraction reaches 90°?

### ray-refraction-3e.iwp

The red line at y = 0 represents a boundary between different media. The path of a light ray is shown in blue. Playing the applet forward or backwards will change the angle of incidence. Which ray is the incident ray? Which is refracted? reflected? In which medium is the speed of light greater? Suppose wavefronts were added to the rays. For which two rays would the wavelength be the same? How would the wavelengths in the 2 media compare? Why?

### ray-refraction-3f.iwp

The blue and gray areas represent different media. Incident, reflected, and refracted light rays in the media are shown. In which medium is the speed of light greater? How do you know? Run the applet to increase the angle of incidence. What will happen when the angle of refraction reaches 90°?

### ray-refraction-4c.iwp

The blue, red, and gray areas represent media of different indices of refraction. (The media are indexed 1,2,3 from the bottom up.) The path of a light ray is shown in yellow. The ray is incident from the blue medium. Playing the applet forward or reverse will increase or decrease the initial angle of incidence. The indices of refraction of the blue and gray media are given. The angle of incidence in the blue medium is given as an output. Determine the angle of refraction in the gray medium. (Why isn't it necessary to know the index of refraction or angle of refraction in the red medium?)

### ray-refraction-3f.iwp

The blue and gray areas represent different media. Incident, reflected, and refracted light rays in the media are shown. In which medium is the speed of light greater? How do you know? Run the applet to increase the angle of incidence. What will happen when the angle of refraction reaches 90°?

### ray-refraction-4c.iwp

The blue, red, and gray areas represent media of different indices of refraction. (The media are indexed 1,2,3 from the bottom up.) The path of a light ray is shown in yellow. The ray is incident from the blue medium. Playing the applet forward or reverse will increase or decrease the initial angle of incidence. The indices of refraction of the blue and gray media are given. The angle of incidence in the blue medium is given as an output. Determine the angle of refraction in the gray medium. (Why isn't it necessary to know the index of refraction or angle of refraction in the red medium?)

### ray-refraction-4e.iwp

The blue, red, and gray areas represent media of different indices of refraction. They are indexed 1 to 3 from the bottom up. The path of a light ray incident from the blue medium is shown in yellow. The angle of incidence is given as an output. Playing the applet forward or reverse will increase or decrease the initial angle of incidence. In order to increase the precision with which the incident angle can be read, decrease the Angle Increment. Also change the Starting angle to something very near the angle you're looking for so that you don't have to step through many angles. The problem is to find the relative indices of refraction n1:n2:n3. That is, find the ratios n2/n1 and n3/n2. For good results, measure angles to the nearest tenth of a degree.

### ray-refraction-4c.iwp

The blue, red, and gray areas represent media of different indices of refraction. (The media are indexed 1,2,3 from the bottom up.) The path of a light ray is shown in yellow. The ray is incident from the blue medium. Playing the applet forward or reverse will increase or decrease the initial angle of incidence. The indices of refraction of the blue and gray media are given. The angle of incidence in the blue medium is given as an output. Determine the angle of refraction in the gray medium. (Why isn't it necessary to know the index of refraction or angle of refraction in the red medium?)

### ray-refraction-4e.iwp

The blue, red, and gray areas represent media of different indices of refraction. They are indexed 1 to 3 from the bottom up. The path of a light ray incident from the blue medium is shown in yellow. The angle of incidence is given as an output. Playing the applet forward or reverse will increase or decrease the initial angle of incidence. In order to increase the precision with which the incident angle can be read, decrease the Angle Increment. Also change the Starting angle to something very near the angle you're looking for so that you don't have to step through many angles. The problem is to find the relative indices of refraction n1:n2:n3. That is, find the ratios n2/n1 and n3/n2. For good results, measure angles to the nearest tenth of a degree.

### ray-refraction-4g.iwp

The blue, red, and gray areas represent different media. The path of a light ray incident from the blue medium is shown in yellow. Rank the media according to the speed of light in them. At some value of the angle of incidence in the blue medium, there will be no refracted ray in the gray medium. In that case, what path will the light ray take in the blue and red media? Is there an angle of incidence for which there is no ray in either the gray or red media?

### ray-refraction-4e.iwp

The blue, red, and gray areas represent media of different indices of refraction. They are indexed 1 to 3 from the bottom up. The path of a light ray incident from the blue medium is shown in yellow. The angle of incidence is given as an output. Playing the applet forward or reverse will increase or decrease the initial angle of incidence. In order to increase the precision with which the incident angle can be read, decrease the Angle Increment. Also change the Starting angle to something very near the angle you're looking for so that you don't have to step through many angles. The problem is to find the relative indices of refraction n1:n2:n3. That is, find the ratios n2/n1 and n3/n2. For good results, measure angles to the nearest tenth of a degree.

### ray-refraction-4g.iwp

The blue, red, and gray areas represent different media. The path of a light ray incident from the blue medium is shown in yellow. Rank the media according to the speed of light in them. At some value of the angle of incidence in the blue medium, there will be no refracted ray in the gray medium. In that case, what path will the light ray take in the blue and red media? Is there an angle of incidence for which there is no ray in either the gray or red media?

### refracted-waves-3.iwp

Plane waves of constant frequency move up the screen, crossing from one medium (blue) into another (gray). The wave speed in each medium is different.

### ray-refraction-4g.iwp

The blue, red, and gray areas represent different media. The path of a light ray incident from the blue medium is shown in yellow. Rank the media according to the speed of light in them. At some value of the angle of incidence in the blue medium, there will be no refracted ray in the gray medium. In that case, what path will the light ray take in the blue and red media? Is there an angle of incidence for which there is no ray in either the gray or red media?

### refracted-waves-3.iwp

Plane waves of constant frequency move up the screen, crossing from one medium (blue) into another (gray). The wave speed in each medium is different.

### refracted-waves-4.iwp

Plane waves of constant frequency move up the screen, crossing from one medium (blue) into another (gray). The wave speed in each medium is different.

### refracted-waves-3.iwp

Plane waves of constant frequency move up the screen, crossing from one medium (blue) into another (gray). The wave speed in each medium is different.

### refracted-waves-4.iwp

Plane waves of constant frequency move up the screen, crossing from one medium (blue) into another (gray). The wave speed in each medium is different.

### refracted-waves-5.iwp

Plane waves of constant frequency move up the screen, crossing from one medium (blue) into another (gray). The wave speed decreases in the upper medium. Since the frequency is constant and speed = frequency x wavelength, the wavelength is less in the upper medium. The white arrows indicate the incident and refracted rays. The rays are perpendicular to the wavefronts.

### refracted-waves-4.iwp

Plane waves of constant frequency move up the screen, crossing from one medium (blue) into another (gray). The wave speed in each medium is different.

### refracted-waves-5.iwp

Plane waves of constant frequency move up the screen, crossing from one medium (blue) into another (gray). The wave speed decreases in the upper medium. Since the frequency is constant and speed = frequency x wavelength, the wavelength is less in the upper medium. The white arrows indicate the incident and refracted rays. The rays are perpendicular to the wavefronts.

### refraction-in-box-2.iwp

A ray of light (red) is incident from air on a transparent medium (blue box). The ray enters from the bottom of the screen. The initial angle of incidence is the smallest that it can be in order to result in a refracted ray emerging from the right side of the box. At smaller angles, the ray would be totally reflected internally from the right boundary of the box. The animation can be played in order to increase the angle of incidence.

### refracted-waves-5.iwp

Plane waves of constant frequency move up the screen, crossing from one medium (blue) into another (gray). The wave speed decreases in the upper medium. Since the frequency is constant and speed = frequency x wavelength, the wavelength is less in the upper medium. The white arrows indicate the incident and refracted rays. The rays are perpendicular to the wavefronts.

### refraction-in-box-2.iwp

A ray of light (red) is incident from air on a transparent medium (blue box). The ray enters from the bottom of the screen. The initial angle of incidence is the smallest that it can be in order to result in a refracted ray emerging from the right side of the box. At smaller angles, the ray would be totally reflected internally from the right boundary of the box. The animation can be played in order to increase the angle of incidence.

### refraction-in-box-3.iwp

A ray of light (red) is incident from air on a box filled with water (blue). The ray enters from the bottom of the screen. Play the applet. Explain why the light ray takes the path that it does. What path will the ray take if the angle of incidence of the ray entering the box is greater than 61.3 degrees?