OPTICS I LENSES AND IMAGES

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1 APAS Laboratory Optics I OPTICS I LENSES AND IMAGES If at first you don t succeed try, try again. Then give up- there s no sense in being foolish about it. -W.C. Fields SYNOPSIS: In Optics I you will learn how a lens forms an image. This is the first step in understanding how a telescope works. EQUIPMENT: Optics bench rail with 3 holders, optics equipment stand (lenses L and L2, image screen I, object mount O, flashlight, iris aperture A); object box, and a chicken. NOTE: Optical components are delicate and are easily scratched or damaged. Please handle the components carefully, and avoid touching any optical surfaces. Your instructor will show you how to use the equipment properly. Part I. The Pinhole Camera R Pinhole Image on Screen First we will use the pinhole camera to show how an image is formed. Light is emitted or reflected from each point of the object through the pinhole and onto the screen; and an image is formed. It is important to note that light from the object is emitted or reflected in all directions, but only the rays of light which pass through the pinhole form an image (see diagram). You can think of the pinhole as selecting only the light which will form an image on the screen. From the diagram above you can also see why the image formed by the pinhole camera is upside-down. O Iris A Image Screen I Flashlight dobj = 20 cm dim 2 (0 cm) (30 cm) 3

2 APAS Laboratory 2 Optics I I. Arrange the optical bench as shown on the previous page: Position holder #2 exactly at the 30 cm mark on the rail (the holder has an arrow that points to the ruled markings. Gently tighten the clamp knob. Install the iris aperture A in the holder, orienting it so that the aperture faces squarely down the rail. Clamp the aperture in place. Close the aperture so that the hole in the center is very small. Slightly loosen the clamp of holder #3, and slide it until it is just behind the lens (at about the 35 cm mark). Install the image screen I (with white card facing the aperture) in the holder. Clamp the screen in place. Turn on the flashlight by rotating its handle. Notice that it has a black arrow, cm in length, on its face. This circle will serve as the physical object which we will use in our study of the lens. Insert the flashlight into mount O as shown in the diagram, so that its face is lined up with the mounting rod, and secure it with the clamp. Position holder # exactly at the 0 cm mark and clamp. Install mount O in the holder and clamp it so that the flashlight points directly at the lens. I.2 On the white screen, you'll see an inverted (i.e. upside-down) image of the arrow. Slowly slide the screen holder #3 away from the aperture. Does the image of the arrow become smaller (converging rays of light) or larger (diverging rays)? I.3 Slowly open the aperture, stopping occasionally to look at the image. What effect does the larger pinhole have on the image? In terms of the ray diagram above, describe what is happening. Now fully open the aperture- do you see an image? Why or why not? Part II. Image Formation by a Lens Now that we've seen how pinholes produce an image we shall see how lenses produce an image. In optical terminology, an object is any physical source of light: it may be self-luminous (a lamp or a star), or may simply be a source of reflected light (a tree or a planet). While a pinhole camera allows only some light to illuminate the image screen, a lens allows much more light; however, light from an object is refracted, or bent, when it passes through a lens. Under certain conditions the refracted light may come to a focus to form an image of the actual object. Lens Focal Plane Focal length Rays of light which enter the lens perpendicular to the lens (and parallel to each other- see above) are refracted such that they all intersect at one point, which is known as the focal point. The distance between the lens and the focal point is the focal length, which is a fundamental property of the lens. It is important to note that lenses are symmetrical- there is no front or back side. Thus lenses have two focal points- one in front and one behind, both of which are the same distance from the lens. Also note that rays of light which pass through the center of the lens are not bent. The concept of focal length is very important, so remember it!

3 APAS Laboratory 3 Optics I Now that you understand the concept of focal length, we will learn how a lens produces an image by ray tracing. There are three rules of ray tracing which will allow you to determine where an image is formed: Lens (2) () f Image f (3) () A ray which enters perpendicular to the lens will be bent to pass through the focal point f. (2) A ray which passes through the center of the lens is not bent. (3) A ray which passes through the front focal point f of the lens will leave perpendicular to the lens. This is just rule () in reverse- remember that lenses are symmetric. The intersection of two or more of these rays will define the location and size of the image formed. The separation between the lens and the object is called the object distance d obj. Since the lens is located at the 30 cm mark on the rail, while the object is at 0 cm, the difference between the two settings gives the object distance: 20 cm. Not surprisingly, the separation between the lens and the in-focus image is called the image distance, d im, You can determine this distance by taking the difference between the location of the image screen (from the rail markings) and the location of the lens (30 mm). Note that distances are always given in terms of how far things are from the lens. Magnification refers to how many times larger the image appears to be compared to the actual size of the object; that is, magnification is the ratio of the image size to object size: Magnification = Image Size Size. (Equation ) f f II. Ray trace the diagram above. Use a ruler. Measure the object distance d obj, the image distance d im and the focal length f. Also calculate the magnification. Does the size of the object affect where the image is formed? f f II.3 Ray trace the diagram above. The object distance has not been changed but the focal length of the lens has been lengthened. Measure d obj and d im and calculate the

4 APAS Laboratory 4 Optics I magnification. How did increasing the focal length change the magnification? Now that you've gotten a feel for how an image is formed try it with the optical bench. O Lens L Image Screen I Flashlight d obj = 20 cm 2 (0 cm) (30 cm) 3 d image II.4 II.5 II.6 Arrange the optical bench as shown above: Replace the iris aperture A with lens L in position holder #2, making sure it is at the 30 cm mark on the rail. Gently tighten the clamp knob. Slightly loosen the clamp of holder #3 (which should be holding the image screen I), and slide it until it is just behind the lens (at about the 35 cm mark). On the white screen, you'll see a bright circular blob that is the light from the object that passes through the lens. Slowly slide the screen holder #3 away from the lens. Does the spot of light become smaller (converging rays of light) or larger (diverging rays)? At some point, you'll notice that the light beam coalesces from a fuzzy blob into a sharp image of the object. At this point tighten the clamp of holder #3. Measure d obj and d im and calculate the magnification. Why is the image formed by the lens brighter than one produced by a pinhole? Unlike the pinhole camera a lens produces an image at a set distance from the lens. We can understand this by looking at ray tracing diagrams. An image is formed only where the rays converge. II.7 Now replace lens L with Lens L2 and slowly move screen holder #3 away from the lens until you get an in-focus image. Measure d obj and d im and calculate the magnification. Compare to the magnification of lens L. Lens L2 has a longer focal length than L. Do your results agree or disagree with your results from the ray-tracing diagram (i.e. does a longer focal-length lens produce a higher magnification)? Part III. The Lens Equation By now you should be convinced that the image location is uniquely determined by the focal length of the lens and the location of the object. This relationship can be written mathematically as the lens equation and is given by: f = d obj + d im. (Equation 3) III. Use your values for d im and d obj from problem II. to calculate f for lens L. Does this agree with your measured value for f? Repeat with the values from II.7 to calculate f for lens L2.

5 APAS Laboratory 5 Optics I HELPFUL HINT: When using your calculator with the lens formula, it is far simpler to deal with reciprocals of numbers rather than with the numbers themselves. Look for the calculator reciprocal key labeled /x. To compute 2 + 4, hit the keys 2 /x + 4 /x =. You should get the result 0.75 ( = 3/4 ). Now if /f equals this quantity, just hit /x again to get f = 0.75 =.33. III.2 In Part II. you encountered a situation where the image was about the same size as the object (magnification =.0). According to the lens equations, the object distance must be how many times larger than the focal length for this special condition to occur? Explain how you arrived at your answer. Does this agree with the ray tracing diagram? Now let s look at a special case that is very important to astronomy. When the object distance is extremely large, then (/d obj ) is extremely small, and for practical purposes can be said to equal zero. For this special case the lens equation then simplifies to: f = d im + 0 or merely f = d im. In other words, when we look at an object very far away the image is formed at the focal planewhich is the plane one focal length away from the lens. Since everything we look at in astronomy is very far away, images of astronomical objects are always formed at the focal plane! You can imagine what is happening by thinking about a ray tracing diagram. Since the object is so far away, all the rays of light coming from the object are coming into the lens parallel to each other and perpendicular to the lens. Thus each ray of light will obey rule # of ray tracing and will pass through the focal point- just like the figure at the start of Part II. We will now investigate this situation by looking at an object that is far enough away that we can assume that f = d im (in photographer's terminology the object is said to be at infinity. ) III.3 Remove the flashlight and mount from holder #. Position the large object box about 3 meters (0 feet) in front of the lens. Focus the screen. Your arrangement should look like this: Large Box Lens L2 Focal Plane I d d obj 300 cm image f III.4 2 Is the image distance approximately equal to the focal length of the lens? 3

6 APAS Laboratory 6 Optics I Part IV. Lens Diameter and Focal Ratio A second important property of a lens is the size of its aperture, or diameter D. IV. Place iris aperture A in holder #. Slide the aperture holder until it is just in front of the lens. Your arrangement should look like this: Large Box Iris Aperture A Lens L2 Focal Plane I Effective d obj 300 cm Diameter D d image f 2 3 If you close, or stop down the iris, you effectively reduce the diameter D of the lens (that is, you reduce the portion of the lens that actually sees the light). III.2 III.3 Analyze the effect of reducing the aperture: As you slowly stop down the lens, what happens to the brightness of the image? Does the image distance change? That is, does the focal plane (where the in-focus image is formed) shift when the lens diameter is decreased? III.4 Does the image size shrink with smaller aperture? Does a portion of the image get "chopped off" because of the smaller lens opening? Although you may be surprised at these findings, they merely point out the fact that each tiny portion of a lens forms a complete image; the final observed image is simply the sum of all of the independent contributions! The brightness of an image produced by a lens (the amount of light per unit area) is actually determined by the ratio of the focal length of the lens to its effective diameter, rather than just the lens diameter alone. This ratio is called the focal ratio, or simply the f-ratio, of the lens: f-ratio = f D. (Equation 4) III.5 As you stop down the aperture, are you increasing or decreasing the f-ratio of the lens? Do larger f-ratios produce brighter or darker images? The f-ratio of a lens is a very important property of a lens because it determines the brightness of the image. Two lenses with the same f-ratio will produce an image of the same brightness (if not the same size).