Optics and Ray Diagrams lab

Optics and Ray Diagrams Ray Diagrams Before you start your ray diagrams, watch the following video that I have made explaining how to do a ray diagram. You should also print out the note from blackboard on Ray Diagrams (under the Lab Notes heading). This has all the definitions that I will describe on the video. How to do a Ray Diagram: https://youtu.be/jA6DSa7Doa0 For the following two problems, you will be creating a ray diagram for a lens with the following characteristics: f = 3 cm diameter = 8 cm Object height (h o ) = 3 cm Problem 1: Object distance = 12 cm Problem 2: Object distance = 8 cm For the next two ray diagrams, you will be creating a ray diagram for a lens with the following characteristics: f = 5 cm diameter = 8 cm Object height (h o ) = 3 cm

Problem 3: Object distance = 12 cm Problem 4: Object distance = 3 cm Some of you may need help to complete problem 4, I have recorded the following video to help you with this so just click the following link. How to do a Ray Diagram for a Virtual Image: https://youtu.be/5TN1sRotyCU We have two types of images that can be created by a lens, a real image and a virtual image. A real image is an image that can be projected onto a screen. A real image is almost always on the opposite side of the lens and inverted when compared to the object. A virtual image is an image that appears to come from a point but is not actually there. This image almost always on the same side of the lens and erect when compared to the object. Using the images formed by your ray diagrams fill the information asked for in the following table. Problem Real or Virtual Inverted or Erect Image Size (cm) Magnification = h i /h o 1 Real inverted 0.8 cm 0.267 2 Real inverted 1.9 cm 0.633 3 Real inverted 1.7 cm 0.567 4 Real inverted 1.875 cm -0.625 Questions 1) What is the relationship between the object distance from the lens and the image size? - The relationship between the object distance and image size depends on the type of lens and the position of the object relative to the lens. Converging lenses can form both real and virtual images, and the image size varies based on the object distance. Diverging lenses always form virtual, reduced-size images.

3) What type of image is produced by the objective? By this I mean real and inverted, or virtual and erect. - When the object is placed beyond the focal length of a converging lens (like the objective lens in this case), the lens converges the incoming light rays to form a real image on the opposite side of the lens. The image will be inverted, meaning it is upside down compared to the object. In summary, when the object is beyond the focal length of the objective lens, the image formed will be real and inverted, regardless of the object's height. 4) What type of image is produced by the eyepiece? - A virtual image is an image that appears to be in a location where light does not actually reach. It cannot be projected onto a screen. In the context of eyepieces, the eyepiece creates a virtual image that appears to be at a certain distance from the eyepiece. This virtual image is what the observer perceives when looking through the eyepiece. So, the image produced by the eyepiece is virtual, meaning it's an image that the observer sees through the eyepiece, and it is magnified to enhance the details of the observed object. 5) What type of image is produced by the telescope? - The type of image produced by the telescope can be real and inverted or virtual and erect, depending on the specific design and components of the telescope. Astronomical telescopes commonly produce inverted images, as the orientation of celestial objects is not relevant for astronomical observations. Terrestrial telescopes, on the other hand, often include additional optics to produce virtual and erect images for convenient terrestrial viewing. 6) Using the equation for magnification in the previous section, calculate the magnification of the objective lens. - M objective = − Imageheight ( h 1 ) Objectheight ( h )

- m objective = − 2.7 cm 3 cm = -0.9 - The negative sign indicates that the image is inverted compared to the object, which is a common convention in optics. So, the magnification of the objective lens is mobjective= -0.9 7) Calculate the magnification of the eyepiece? - Magnification of the eyepiece is approximately 1.111 m objective = − 2.7 cm 3 cm = -0.9 −1 = (−0.9) × m eyepiece m eyepiece = −− 1 − 0.9 m eyepiece ≈ 1.111 8) Knowing that the image seen through the eyepiece is the image of the telescope, calculate the magnification of the telescope. - m total= m objective × m eyepiece - m total = ( −0.9) × (1.111) - m total ≈ −1 - The negative sign indicates that the final image is inverted. The total magnification of the telescope is approximately -1. This means that the image seen through the telescope appears inverted and at the same size as the original object, which is a common configuration for astronomical telescopes . 9) According to theory, if we multiply the magnification of the objective by the magnification of the eyepiece we should get the magnification of the telescope. Do this calculation to prove that this is true. - m total = m objective × m eyepiece - mtotal = (−0.9) × (1.111) - m total ≈ −1

These two images are of two converging lenses. The one on the left has a focal length of 10 cm; the one on the right has a focal length of 20 cm. In both images, the object is placed 30 cm away from the lens. 10) What is the relationship between the brightness of the image and the focal length of the lens? - The relationship between the brightness of the image and the focal length of the lens is indirect. It is primarily determined by the size of the lens's aperture. A larger aperture allows more light to pass through, resulting in a brighter image, while a smaller aperture restricts the amount of light, leading to a dimmer image. The focal length itself does not directly influence the brightness; it is the aperture size that plays a key role in determining the brightness of the formed image.

These two images show a 10 cm focal converging lens. The photo on the right shows the image resulting from the lens with its full aperture. The image on the left shows the same lens, but with an image formed using a 3 cm aperture. 11) What is the relationship between image brightness and the aperture of the lens? - The relationship between image brightness and the aperture of the lens is straightforward: a larger aperture allows more light, leading to a brighter image, while a smaller aperture restricts light, resulting in a dimmer image. Photographers often use different aperture sizes (expressed as f-numbers) to control the exposure and depth of field in their photographs, understanding the impact of aperture on image brightness is fundamental to capturing well- exposed images. If we were to build a telescope with two converging lenses, the magnification is given by: ? 𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑀= ? 𝑒𝑦𝑒𝑝𝑖𝑒𝑐𝑒 We have four lenses: Lens Focal Length (cm) A 4 B 3.5