Dispersion & Thin Lenses PhET Lab
Topics:
- Applications of Refraction and Dispersion
- Rainbows
- Spectroscopy
- Chromatic Aberration
- Lenses
- Go to the PHET physics simulations, and open the “Bending Light” simulation.
- Click “Prisms”.
- Click the red button on the laser.
- Keep the laser set to a single ray and red (650 nm wavelength) light. For the Environment material, keep it set to Air.
- For the Object material, choose Water from the drop down list.
- Drag a sphere from the container to be aligned as shown below.
- Click the Reflections checkbox, and make sure the Normal checkbox is also checked.
- Notice that there is an incident beam coming from the laser which gets partially reflected off the front side, the rest is refracted into the water sphere, then partially reflected off the inside of the water sphere, and the rest is refracted back out to the air. The reflected beam then travels through the water sphere, and is refracted into the air. And there is another faint reflected beam, etc.
- Notice that for the reflected beam, the angle of incidence equals the angle of reflection, which you can see because the angles are symmetric about the normal dashed lines. Also, you can measure it with the protractor if you like.
- Now click off the Normal setting and on the multiple beams setting. Try different combinations of Object materials and Environment materials and Object shapes. How does the amount by which the light is bent change with the index of refraction of the Object? How does the amount by which the light is bent change with the index of refraction of the Environment? Notice that what matters is the relative indices of refraction of the Environment and the Object.
- Divergence occurs when the light beams bend outwards from one another (i.e. the sphere is acting as a diverging lens). Convergence is when the light beams are bent towards one another (i.e. the sphere is acting as a converging lens).
- If the index of refraction of the Environment is less than the index of refraction of the Object, the light is______ in going through the sphere.
- diverged
- not bent
- converged
- If the index of refraction of the Environment is equal to the index of refraction of the Object, the light is______ in going through the sphere.
- diverged
- not bent
- converged
- If the index of refraction of the Environment is greater than the index of refraction of the Object, the light is______ in going through the sphere.
- diverged
- not bent
- converged
- Now set up your simulation as shown below. Environment: Air. Light: White. Object: Water Sphere. Reflections: on.
- Notice the dispersion. Dispersion (meaning “to distribute over a wide area”) is the phenomenon of different wavelengths of light bending differently when transitioning between mediums (due to the index of refraction changing slightly with wavelength), and thus spreading out into a spectrum all of the wavelengths that make up a light beam.
- In refraction, which leads to dispersion, which color of light is bent most?
- Red
- Yellow
- Green
- Blue
- In refraction, which leads to dispersion, which color of light is bent least?
- Red
- Yellow
- Green
- Blue
- Dispersion is the basis of rainbows, and double rainbows.
- Dispersion is the basis of spectroscopy, where electromagnetic radiation from a source is passed through a prism, and the resulting spectrum is analyzed to give information about the source.
- Dispersion is the source of chromatic aberration, which is a downside of refracted images where not all of the wavelengths of light focus at the same point, which causes halos of color and blur. This can be present in any devices (cameras, telescopes, microscopes, etc.) that use lenses to focus light.
- Given what you’ve seen with the phenomenon of dispersion, which color should be at the bottom of a rainbow?
- Red
- Yellow
- Green
- Blue
- Violet
- Given what you’ve seen with the phenomenon of dispersion, which color should be at the top of a rainbow?
- Red
- Yellow
- Green
- Blue
- Violet
- Refraction is the basis of our use of lenses. There are two types of lenses. Converging lenses bend light inward, and diverging lenses bend light outward. Converging lenses are convex and have a positive focal length, and diverging lenses are concave and have a negative focal length.
- With lenses, there is an object that is the source of light, and an image of that object that is formed by the lens. Images that are formed with actual light rays are called real images, and can be projected onto a screen/flat surface for viewing. Images that are formed with projections of light rays are called virtual images, and cannot be projected onto a screen/flat surface for viewing.
- Open the PHET: Geometric Optics:
- Set up your screen as follows:
- Checked: Many Rays, Virtual Image, Ruler
- Curvature Radius: 0.8 (this is a spherical lens, and this sets the radius of curvature)
- Refractive Index: 1.53
- Diameter: 1.3
- The pencil on the left is the object. The light rays shown are those that originate from the tip of the pencil. The pencil on the right is the image of the object formed by the light rays, which pass through lens. The object distance, o, is the distance from the object to the center of the lens. The image distance, i, is the distance from the image to the center of the lens. The focal length, f, is the distance from the center of the lens to the X on either side.
- Pencils don’t emit light. However, if you are in a room with a light on, you can see the tip of a pencil, because light rays from the light source in the room reflect off of the tip of the pencil and into your eyes. These reflected rays from the tip of the pencil are the ones shown in the simulation.
- What type of lens is this spherical lens? (Select all that apply)
- Converging
- Diverging
- Positive (a positive lens has a positive focal length)
- Negative (a negative lens has a negative focal length)
- Concave (biconcave)
- Convex (biconvex)
- Using the ruler, how far are the focal lengths from the center of the lens?
- 40 cm
- 60 cm
- 75 cm
- 85 cm
- For an object at a distance greater than the focal length of a convex lens, the image is (use the thin lens equation with f = 10, and o = 20, solve for the image distance and the magnification, and do the ray tracing using the principal rays as shown in lecture, then answer the question below)
- Upright
- Inverted
- On the opposite side of the lens
- On the same side of the lens
- Formed with actual light rays
- Formed with projections of light rays, but not actual light rays
- Virtual
- Real
- Using the mouse, slowly slide the pencil (object) horizontally toward the lens, but do not slide it past the focal point of the lens (the “X” on the line). Observe what happens to the image.
- As an object at a distance beyond the focal point nears the focal length of a concave lens, the image gets (Select all that apply) (use the thin lens equation with f = 10, and o = 30 and 20 respectively, solve for the image distance and the magnification, then answer the question below)
- farther from the lens
- closer to the lens
- smaller
- larger
- Using the equation for magnification, at what distance for the object do the object and image appear to be the same size? How do these distances relate to the focal length, f, of the lens?
- When an object is located at ___, the image will be at a distance of ___ from the lens and the same size. (use the thin lens equation to solve this problem)
- 3f, f
- f, 3f
- 2f, 2f
- f, f
- Now solve the thin lens equation for an object directly centered on the focal point (i.e. object distance = f). What is the image distance for this scenario? What does this imply about the light rays that emerge from the other side of the lens? (see the possible answers for 38 and 39 to think through the last two parts of this question)
- When an object is on the focal point of a lens, the light rays emerging from the lens …
- converge
- are parallel
- diverge
- none of the above
- What does that fact indicate about the image?
- The image is located at the focal length on the other side.
- The image is located in the lens.
- The image is located at infinity.
- None of the above.
- Now slowly slide the object horizontally past the focal point toward the lens until you get almost to the lens. Observe what happens to the image. Observe what makes up this image.
- As an object at a distance less than the focal length nears the surface of a convex lens, the image gets (Select all that apply) (use the thin lens equation with f = 10, and o = 7 and 3 respectively, solve for the image distance and the magnification, then answer the questions below)
- farther from the lens
- closer to the lens
- smaller
- larger
- For an object at a distance less than the focal length of a convex lens, the image is
- Upright
- Inverted
- On the opposite side of the lens
- On the same side of the lens
- Formed with actual light rays
- Formed with projections of light rays, but not actual light rays
- Virtual
- Real
- Now, by measuring with the on-screen ruler and by moving the object, fill out the table below. Image distances to the right side of the lens should be treated as positive (i.e. on opposite side as object), and image distances to the left side of the lens should be treated as negative (i.e. on same side as object). Make sure the index of refraction is set to 1.53.
Complete the table below:
| Focal length, f | Object distance, o | Image distance, i | 1/f | 1/o | 1/i |
| 40 | 30 | -120 | |||
| 40 | 50 | 200 | |||
| 40 | 120 | 60 | |||
| 40 | 150 | 54.54 | |||
| 40 | 200 | 50 |
- Do you see a relationship between any of the columns in the table above, i.e. what is an equation that relates the variables, f, o, and i? This equation is called the thin lens equation.
- Return the pencil to its original position, and unclick many rays, and click principal rays. Unclick the ruler. Notice that (1) the top ray that travels from the pencil tip into the lens parallel, always goes out through the focal point on the other side, (2) that the middle ray always goes from the tip of the pencil through the center of the lens, and (3) that the bottom ray always goes through the focal point on the same side as the object (or projects from it) and goes out parallel on the other side. Notice that these three rays which originate at the object tip, always intersect at the image tip. These three rays are called the principal rays and can be drawn to figure out where the image will be and what size it will be.
- Note that some times the image is larger, and others the image is smaller. When |i|<o, is the image larger, smaller, or the same size? (use the equation for magnification shown in lecture to solve the next few problems)
- When |i|>o, is the image larger, smaller, or the same size?
- When |i|=o, the image is larger, smaller, or the same size?
- Given the above, we can find that the magnification (i.e. height of image = – height of object * M). If , then the image is inverted. If , then the image is smaller. If , then the image is larger. If , then the image is the same size.
- Finally, turn Many Rays back on and Principal Rays off. Now change the index of refraction. How does changing the index of refraction change the focal length? How does changing the index of refraction change the location of the image?
- If o
- Decreases f
- Increases f
- Decreases the magnitude of i
- Increases the magnitude of i
- If o>f, increasing the index of refraction of the lens (select all that apply)
- Decreases f
- Increases f
- Decreases the magnitude of i
- Increases the magnitude of i
- Given how the index of refraction affects the location of the image, would the thin lens equation hold true for lenses of all materials? What is the index of refraction of glass? (look up n for glass in Google)
