Geometric Optics: Reflection and Refraction, Lecture notes of Optics

Download Geometric Optics: Reflection and Refraction and more Lecture notes Optics in PDF only on Docsity! Lab 11: Geometric Optics Part 1: Reflection and Refraction With this experiment we begin the study of light and its interaction with various optical components. In this experiment we shall study the limit where the wave nature of light can be neglected and we can consider a train of light waves to be a straight line, or ray. Wave effects are usually unimportant when the size of the optical component is much larger than the wavelength of the light which is interacting with it. Situations where the wave nature of light must be taken into account will be studied in a later experiment. In order to define terms, Figure 1 shows a light ray (in an optical medium having index of refraction n1) incident from the left onto an interface separating the first optical medium from a second optical medium having index of refraction n2. The index of refraction of an optical medium is defined as the ratio of the speed of light in a vacuum (symbolized by “c”) to the speed of light in the optical medium: n = c/v Equation 1. In general, the speed of light in an optical medium depends on the wavelength of the light. Therefore the index of refraction will also depend on the wavelength. This dependence of index of refraction on wavelength, called dispersion, explains how a prism is able to spread a beam of white light into its component colors. As the figure shows, an incident ray of light will, in general, split into two parts as it hits the surface separating the two media. One part, the reflected ray, will remain in the first medium, and will make an angle (measured relative to the normal to the surface) that equals the angle the incident ray made with the normal. Expressed mathematically, the law of reflection says: θι = θι' Equation 2. 1 The second part, the refracted, or transmitted, ray, proceeds on into the second medium. This ray makes an angle, θ2, measured relative to the normal line in the second medium, which is determined by the Law of Refraction (also known as Snell’s Law). Snell’s Law states that: n1 sin θ1 = n2 sin θ2 Equation 3. Note: Figure 1 (previous page) represents a situation where n2 > n1, resulting in θ2 being smaller than θ1. If n2 < n1 the situation is reversed and θ2 becomes greater than θ1. This would be the case, for example, if a ray of light already in an optical medium such as glass or transparent plastic was exiting back into an optical medium such as air. Figure 2 depicts this situation: Note that this means that if θ1 is allowed to get larger, θ2 will eventually approach 90o. But when θ2 = 90o it means that no refracted ray goes into the second medium. Rather the incident ray is completely reflected back into the first medium. This phenomenon is called total internal reflection. Clearly this can happen only when a ray is incident from a medium of higher index of refraction into a medium of lower index of refraction. Referring to Equation 3, we see that total internal reflection first occurs when sin θ2 = 1.0. When this happens, θ1 is said to be at its critical value, θcrit. This critical value of θ1 can therefore be found by setting sin θ2 = 1 in Equation 3 and solving for sin θ1. Remembering that θ1 is now θcrit, the resulting equation is: sin θcrit = n2/n1 Equation 4. Procedure: Part 1a: Attach the optical table to the optics bench and place the semicircular plastic lens on the optical table. The center of the flat face of the lens should be aligned with the line on the table labeled “component”, and the normal line should intersect the center of the flat face. See Figure 3 on the next page. 2 can be used to determine the focal length, (f) of the lens. The focal length is a measure of the imaging power of the lens. In this part of the experiment we will study how a lens produces an image. Ray Tracing: A simple way to understand how a lens works is to follow a procedure called ray tracing. A ray is an imaginary straight line that traces the path of the light. By following the paths of two or three specially selected rays coming from the same point on the object, it is possible to construct a diagram which visually depicts the creation of the image point of that point on the object. As they pass through the lens, the rays will be refracted at both the front and rear surfaces of the lens. After passing through the lens, the rays will, in general, intersect at some point in space. This point is the image point we are looking for. If the lens is not too thick, it is possible to simplify the ray tracing process by assuming that the ray changes direction not at the surfaces of the lens, but rather at a plane through the middle of the lens. If the lens is sufficiently thin (such at the lenses we will be using in the lab), this approximation does not introduce and errors. Your textbook gives a discussion of ray tracing, along with several examples. Be sure to review this information before coming to lab. Be sure that you clearly understand how to construct accurate ray diagrams for both convex (positive) and concave (negative) lenses for all of the categories of images these lenses are capable of producing. 2a: Focal Length of a convex lens For thin lenses, object distance, p, image distance, i, and focal length, f, are related by a simple equation: 1/p + 1/i = 1/f Equation 5 The ratio of the height of the image, hI, to the height of the object, hO, is called the transverse magnification, m. It can be calculated from: hI /hO = m = -i/p Equation 6 You will now make measurements involving the focal length of a positive lens by three different methods, each method intended not just to gather data and do a calculation, but also to familiarize you with the characteristics of the image produced in each case. Method A: If the object is far from the lens, the value of 1/p becomes so small that 1/i approximately equals 1/f. Or, in other words, i approximately equals f. The image produced will, in general, be quite small and will be located approximately at the focal point of the lens. 5 Procedure: 1. Remove the light source table from the optics bench and mount the lens and viewscreen on the optical bench.. 2. Orient the bench so that the image of a distant object (a lightbulb at the far end of the lab, for example) appears on the screen. 3. Adjust the location of the screen until the image is sharp and clear. Measure and record the distance between the lens and the screen. This image distance approximately equals the lens’s focal length. 4. Draw a ray diagram depicting how this image is formed on the screen. Method B: When the object distance is twice the focal length of the lens, Equation 5 can be solved to show that the image distance is also twice the focal length. The transverse magnification is then -1. This means that the image is the same size as the object, but is inverted. Procedure: 1. Solve Equation 5 and show that if p = 2f then i = 2f also. 2. Now mount the light source near one end of the optical bench and orient the source so that the face with the crossed arrows on it points toward the lens. Using your approximate focal length from Method A as a guide, locate the screen approximately 4f from the light source, and place the lens halfway between them. 3. Carefully adjust the positions of the lens and the screen until you have a clear image that is exactly the same size as the object (use the vernier caliper at your lab station to measure the size of the image and the size of the object), and so that the object distance equals the image distance. 4. Measure and record your values for object distance and image distance and use them to calculate the focal length of your lens. 5. Draw an accurate ray diagram depicting how the image is formed in this case. 6 Method C: For an arbitrarily chosen object distance, and a lens of known focal length, Equation 5 can be used to predict the location of the image. Your results from Method B should give you an accurate and reliable value for the focal length of your lens. There are three possible ranges in which the object distance can fall: p > 2f, f < p < 2f, and p < f. For each of these ranges, do the following: Procedure: 1. Choose an object distance and use Equation 5 to calculate a predicted value for the corresponding image distance. 2. Setup your optics bench using your chosen value of object distance and your calculated value of image distance and see whether you do indeed get an image at the predicted location. If so, measure and record your experimental values of p and i and compare the predicted and experimental values of the image distance. 3. Draw an accurate ray diagram depicting the formation of the image in each case. 7 Part 2: Image Formation by Lenses Method A: Approximate focal length__________________ Ray Diagram: Method B: Solve Equation 5: Object Distance______________________ Image Distance_____________________- Calculated focal length:________________ Ray Diagram: 10 Method C: 1. p > 2f: Object distance________________________ Predicted image distance________________ Measured image distance________________ Ray diagram: 2. f < p < 2f: Object distance_________________________ Predicted image distance_________________ Measured image distance_________________ Ray diagram: 3. p < f: Object distance__________________________ Predicted image distance__________________ Measured image distance__________________ Ray diagram: 11 Conclusion: Summarize your understanding of image formation by lenses by completing the following table.. Under what circumstances is the image larger than the object? Smaller? Same size? Are there circumstances when no (real) image forms at all? What about relative image size and orientation? Object Distance Image Size? (magnified, diminished) Image Orientation? (upright,inverted) Image? (real,inverted) p > 2f f < p <2f p < f p = infinity 12