PHYS 202 Notes, Week 9

Transcription

1 PHYS 202 Notes, Week 9 Greg Christian March 22 & 24, 206 Last updated: 03/24/206 at 2:23:56 This week we learn about images by mirrors, refraction, and thin lenses. Images Spherical Mirrors First let s consider the formation of an image from light reflecting off a concave (curved inward) spherical mirror. A picture of this is shown in Figure. Before moving on, let s define some terms that dictate the geometry of the mirror:. Center of curvature, C. This would be the center of the sphere if the mirror were extended to a complete sphere. 2. Radius of curvature, R. This is the radius that the mirror would have if it was extended into a complete sphere. Here the radius of curvature is positive because the outgoing light is on the same side of the mirror as the center of curvature. 3. The vertex, V. The center of the mirror surface. 4. Object point, P. The location of the object whose image is being formed. 5. Optic axis. The line connecting P, C, and V. For the moment, assume PV is greater than R. 6. Object distance s. The distance from the object point to the vertex. 7. Image distance s. The distance from the image point to the vertex. Important points Spherical (convex or concave) mirrors form images that are magnified and at a different location than the originals. Images for concave mirrors are real, while convex-mirror images are virtual. Important equations Spherical mirror Magnification /s + /s = /f m = y /y f = R/2 = s /s One way to remember convex vs. concave is that concave mirrors form a cave, or hollowed out surface, by curving inward. Armed with this description, we can discuss the image formation at point P. Let s consider a ray PB, at angle α with the optic axis, striking the mirror at point B in the figure. The angles of incidence and reflection are both θ. It turns out that as long as α is small, all such rays intersect at the same point P, regardless of the angle α. The concept above can be proven algebraically, as shown in the textbook. Here we ll just present the main result, describing the relationship between the image distance s, object distance s, and radius of curvature R: s + s = 2 R. () As you can see, the relationship is completely independent of the angle α. Note that this is an approximate equation, since it makes the assumption that α is small. The term for this type of approximation is paraxial approximation.

2 phys 202 notes, week 9 2 Because the light rays really do converge at point P, this is an example of a real image. Contrast with the virtual image formed by a plane mirror, which only had apparent convergence behind the mirror. When dealing with spherical mirrors, we can define something called the focal point, F, which is the image point formed when the object is infinitely far away from the mirror, i.e. when s. We call the distance to the focal point the focal length, f. This can be derived from Eq. (): Figure : Image formation from a concave spherical mirror. + f = 2 R (2) f = R 2. (3) Note that f > 0 for a concave mirror. For a convex mirror, f < 0. Equation () is usually written in terms of the focal length f rather than the radius R. In this form, it becomes s + s = f. (4) Magnification As shown in Figure 2, the height of a finite object is not the same at the object point (point P) as at the image point (point P ). This leads to the phenomenon of magnification. Assuming some object has height y at the object point and Y at the image point, we can define magnification,

3 phys 202 notes, week 9 3 m, as m = y y (5) = s s. (6) Note also that the image is inverted in Figure 2: the direction of the height at the image point is pointing downward, as opposed to upward as it is at the object point. This follows naturally from Eq. (6): since the object point and image point are both on the same side of the vertex, they both have the same sign and m is a negative number. More generally, Images with a positive magnification, m > 0 are not inverted. Images with a negative magnification, m < 0 are inverted. Figure 2: Magnification of an image due to a concave mirror. Also note that despite the name magnification, the absolute value of m can be either greater than or less than unity. This means the magnified image can either be smaller or larger than the original: If m >, the image is larger than the original. If m <, the image is smaller than the original. If m =, the image is the same size as the original. To summarize, the sign of the magnification tells whether the object is upright or inverted, and the magnitude tells whether it s smaller or larger (or the same) than the original. Convex mirrors Figures 3 and 4 show an analysis of a convex mirror. In general, the situation is the same as for concave mirrors, with Eqns. () and (6) describing the image point location and magnification, respectively. However, there are some notable differences: Figure 3: A convex mirror.

4 phys 202 notes, week 9 4 The radius of curvature R is negative. The light rays are not reflected through the focal point F. Instead they look as if they had come from a point f behind the mirror. Following the point above, convex mirrors form a virtual, rather than a real image. In general, we can use Eqns. () and (6) without regard to whether the mirror is convex or concave, as long as we treat the signs of the different quantities correctly and consistently. Figure 4: Focal point of a convex mirror. Graphical Methods for Mirrors Figure 5 summarizes a graphical method for finding the image point and image size reflected by a spherical mirror. It involves drawing Figure 5: Principal rays in convex and concave mirrors.

5 phys 202 notes, week 9 5 four principal rays which emanate from a point on the object (which cannot be on the optic axis). The properties of these four rays are also summarized in the figure. All of the rays pass through the pioint Q, which corresponds to the image version of their common origin point Q. Once these have been found, the path of any other ray can easily be deduced simply be ensuring that it passes through the point Q. Lenses Refraction at a Spherical Surface Figure 6 shows the situation when light is refracted by a spherical medium boundary, with index of refraction n a where the light originates and n b past the refraction bnoundary. As with mirrors, we analyze things in terms of the radius of curvature R. It turns out that for a small incident angle α, the image point is again independent of α, just like with mirrors. The equation describing the relationship between the image point and the object point is n a s + n b s = n b n a. (9) R We can also derive an equation for the magnification due to spherical refraction: m = n a s (0) n b s As with mirrors, the radius of curvature R can either be positive or negative. However, there is one subtle difference: we define the radius of curvature relative to the direction of the outgoing light. Thus the image in Figure 6 has a positive radius of curvature because the surface curves towards the direction of outgoing light. However, the geometry of the problem is the opposite for mirrors: it looks like a convex mirror which would have a negative radius of curvature. The reason for this, of course, is that mirrors reflect light, changing its direction, while the refracting surface transmits light. A special case to consider is a plane refracting surface. In this case, R in Eq. (9), and n a/s + n b/s = 0. Plugging this into Eq. (0) gives the result for the magnification: Important points Spherical refracting surfaces can be analyzed in terms of the radius of curvature, similar to spherical mirrors. The same equations can be used to analyze lenses as mirrors, but you nust take care to pay attention to the signs of quantities. Important equations Spherical refraction image point n a/s + n b/s = (n b n a)/r Spherical refraction magnification m = n as /n b s Thin lens equation /f = /s + /s (7) = (n ) (/R /R 2 ) (8) m =. () In other words, a plane surface produces no magnification: the image is the same size as the object. Figure 6: Refraction at a spherical surface.

6 phys 202 notes, week 9 6 Thin lenses One commonly used optical instrument is a lens, which consists of two refracting surfaces. Here we ll discuss the simplest case, a thin lens, which has two spherical surfaces close enough together that the distance between them is negligible. An analysis of a thin lens is shown in Figures 7 and 8. There s a few important characteristics to be aware of:. The lens has two focal points, F and F 2 : F 2 is the point where incoming parallel rays converge (top of Figure 7). All incoming rays passing through F will result in parallel outgoing rays (bottom of Figure 7). 2. The distance from the center of the lens to either F or F 2 is the same, and is called the focal length, f (top of Figure 7). 3. The lens shown in Figure 7 is called a converging lens because incoming parallel rays all converse at the same real image point after passing through the lens. It s also called a positive lens because f > 0. Figure 8 shows how we can use ray tracing to find the image distance for a thin lens. First, we draw one ray which starts out parallel and then is refracted to pass through F 2. We also draw a ray which passes through the center of the lens, which results in no change in direction since the two surfaces are parallel at the center. From these rays, we can do some trig to get the result: Figure 7: Focal points of a thin lens. i.e. the exact same as for spherical mirrors. s + s = f, (2) Figure 8: Analysis of a thin lens.

7 phys 202 notes, week 9 7 Similarly, the magnification is the same as for mirrors: m = y y = s s. (3) Essentially you can apply the exact same equations, but still must take care to pay attention to the signs of the quantities to get the correct results. In addition to the converging lens already mentioned, we can also deal with diverging lenses. An example is shown in Figure 9. Here incoming parallel rays diverge, or spread out after the lens. In this case, the focal points are flipped relative to the converging lens: F 2 appears in front of the lens, at the point where parallel incident rays look like they re diverging from. F appears behind the lens, at the point there emergent parallel rays would converge if the lens wasn t there. As a result, the focal length of a diverging lens is negative. For this reason, it s sometimes called a negative lens because f < 0. Equations (2) and (3) apply equally well to diverging lenses. However, take care to use the correct signs, in particular to remember that f is a negative quantity for diverging lenses. Thin Lens Equation Figure 9: A diverging thin lens. Figure 0: Illustration of lens geometry used in the lensmaker s equation. So far, we ve seen Eq. (2) describing the image distance in terms of the focal length and object distance. Similar to mirrors, we can also express f in terms of the lens geometry and index of refraction. This is called the lensmaker s equation. As an example, consider the lens shown in Figure 0. The focal length can be expressed as 2 f ( = (n ) ), (4) R R 2 2 See the textbook for a derivation; we won t repeat it here.

8 phys 202 notes, week 9 8 where n is the index of refraction of the lens and the radii R and R 2 are as shown in Figure 0. From this, we can write down what is called the thin lens equation, fully describing the image distance in terms of the lens geometry and object distance: Graphical Methods s + s = (5) f ( = (n ) ). (6) R R 2 Similar to mirrors, we can analyze lenses in terms of principal rays. However, for lenses there s only three principal rays as shown in Figure. These are as follows:. Ray parallel to the axis: Passes through F 2 for a converging lens, or appears to come from F 2 for a diverging lens. 2. Ray through the center of the lens: is not bent; continues along its original path. 3. Ray through, away from, or proceeding towards the first focal point F : emerges parallel to the axis. Figure : Principal rays for lenses.