Lens Equation Purpose

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1 Lens Equation Purpose To verify the lens equation for both a converging lens and a diverging lens. To investigate optical systems. To find the focal lengths of a converging lens and a diverging lens. Background By definition, light can travel through transparent materials. Its speed, though, will be slower than through a vacuum. The ratio of the speed in vacuum to that in the material is called the index of refraction n. When a ray of light strikes the in ter face sur face be tween two me dia at a non-per pen dic u lar an gle, its direction changes. This bend ing of a light ray is called re frac tion. What we are concerned with in this experiment is what happens when light strikes a piece of glass called a lens. The lens is shaped such that it has a section of a spherical surface on one side and a section of a different spherical surface on the other. The light goes from air to glass and back to air. If the air-glass and the glass-air surfaces are parallel, the light ray continues in its original direction. When the spheres defining the sections are different sizes, the light spreads out either immediately from the glass or after light beams converge to an area. There are two types of lenses, converging and diverging. For a converging lens, incoming light rays that are parallel to the optical axis and parallel to each other converge to a common focal point after going through the lens. The distance from the center of the lens to this point is the focal length. For a diverging lens, incoming parallel rays appear to be coming from a focal point behind the lens when viewed from the other side. A fundamental relationship studied in geometrical optics correlates the distance d o be tween an ob ject and a lens, the dis tance d i be tween the lens and the im age po si tion, and the fo cal length f of the lens. (A quick aside: the no men cla ture for the ob ject and im age dis tances var ies widely and what is used here may not be the same as your text. The ideas are the same.) This ba sic re la tion ship can be ex pressed by the equa tion = + (1) f do di known as the thin lens equation. In this experiment, several pairs of object and corresponding image distances are measured to check the func tional form of the equation and to find a value for the focal length of a lens. An example of a converging lens is a magnifying glass. It is thickest at the center of the lens which causes the parallel rays of light to converge to a common focal point. Each lens has two focal points, one on each side of the lens equidistant from the center of

2 the lens. By convention the focal distance f is positive for a converging lens and negative for a diverging lens. Ray diagrams We can represent the path or beam of a light wave with an arrow or ray. By drawing three rays from the top of an object represented by an upright arrow, we can graphically find the image properties resulting from light going through a lens of a particular focal length. Given a known object size and distance, we can determine the size, the position, and the type of image. 1. A ray traveling from the object parallel to the optical axis of the lens is bent by refraction and goes through the focal point on the image side of the lens. 2. A ray going through the focal point on the object side emerges parallel to the axis of the lens. 3. A ray going straight through the center of the lens is not appreciably deviated. Where lines containing these rays intersect is the point of the image arrow. If this point is on the side opposite the object, the image is real, that is, an actual image would appear on a screen at that point. If the intersection point is on the same side as the object, the image is virtual and no image would appear on a screen. See Figs. 1 and 2. d o 1 d i object 3 2 h o α β α β f f h i image (real) lens Figure 1 Ray di a gram for a con verg ing lens where the ob ject dis tance is greater than the fo cal length of the lens.

3 1 h i 2 image (virtual) h o d i α object f d o lens f 3 Figure 2 This is an ex am ple of a con verg ing lens ray di a gram where the ob ject dis tance is less than the fo cal length. For more precision, we use Eq. 1 to find the image distance. The type of image depends on the relationship between the object distance and the focal distance. Derivation of lens equation The magnification M of a lens is the ra tio of the im age size or height to the ob ject height. A neg a tive mag ni fication signifies an in verted im age. By this con ven tion, a pos i tive height in di cates the ar row is erect and a neg a tive height means the ar row is point ing down (val id for both ob ject and im age). From the sim i lar tri an gles with an gle α shown in Figs. 1 and 2, we can re late the magnification to the object and image distances: M = h h i o di = (2) do Using the triangles with angle β in Fig. 1, we can come up with yet an other ex pres sion for the ra tio of the heights: h h i o = d i f f di = (3) d o Dividing the last equation by d i gives = (4) f d d i o and solving for the inverse focal length gives Eq. 1, the lens equation.

4 Diverging Lens A diverging lens is thinner in the center than on the edges, and rays parallel to the axis diverge outward from the lens so that the rays seem to come from a focal point behind the lens. The focal length for a diverging lens is negative, as noted earlier. The image distance is also negative (provided d o is positive), since = (5) d f d i o This means that the image formed by a diverging lens with a real object is always virtual. The magnification is given by Eq.3. Ray diagrams for a diverging lens can be drawn, similar to those for the converging lens. Refer to your text for examples. Finding the focal length experimentally When the focal length of a converging lens is not given, there are two simple ways of finding its value: Images of many objects 1. Find the corresponding image distances for various object distances and plot either 1/d i vs. 1/d o or 1/d o vs. 1/d i. The y-in ter cept for both of these graphs is the neg a tive re cip ro cal of the fo cal length. Image of infinite object 2. Focus on a object far enough away that we can assume that 1/d o is zero com pared to 1/d i. Then the im age dis tance is the fo cal length. Determining the focal length of a diverging lens is more complicated, since it does not produce a real image of a real object. (Again, a real image is one that can be projected on a screen and hence measured) We have to use a converging lens to come up with an image of the lighted object that we can focus on the screen. This is an example of an optical system. The analysis is outlined in the following section and an example is given. Lens combinations When more than one lens is used in an optical system, each lens should be treated individually to find the combined effect. The image of one lens becomes the object of the next lens. If we knew the focal lengths of the lenses and the object distance, we could predict where the image would be, its magnification, whether it was inverted, and whether it would be real or virtual. For example, in the second part of this experiment, both a converging lens and a diverging lens are used to produce an image. We know the focal length of the converging lens and we want to find the focal length of the diverging lens. Suppose that the diverging lens is closer to the object and that there is a focused image on the screen. The known values besides the converging lens focal length f con are the po si tions of the ob ject x obj, the two lenses x div and x con, and the im age x im. First we work back wards to find the ob ject po si tion of the con verg ing lens, then use this as the im age po si tion of the di verg ing lens.

5 1. We easily find the object distance of the diverging lens and the image distance of the converging lens: d = x x o div div obj d = x x i con im con (6) 2. Use the lens equation to find the object distance of the converging lens : d o con = 1 1 fcon d i con 1 (7) 3. Let D=x con -x div be the dis tance be tween the two lenses. The ob ject of the con verg ing lens is the im age of the di verg ing lens. So d = D d (8) i div o con 4. For a single point calculation we could then use the lens equation to find the focal length f div. This is not the best way to do things ex per i men tally. We use the method of Im ages of many ob jects: Find a bunch of val ues of d o div and d i div and plot 1/d o div vs 1/d i div (or vice versa). The y-in ter cept is 1/f div. Procedure You need the following equipment:» optical bench» lighted object» converging lens, 13 cm nominal focal length (far away objects viewed through the lens at arm s length are inverted)» diverging lens, -20 cm nominal focal length» glass screen» optical bench fixtures for the various elements Notes The object position has been set to x obj = 3.50 cm. Keep it there. No, the fix ture is not set at ex actly 3.50; the plas tic slide that is the ob ject is off set from the fix ture read ing. This has been ac counted for in the place ment. The glass plate, the plastic slide that projects the arrows image, and the lens should be perpendicular to the optical bench length. Converging lens: Images of many objects Note and record that the object is at 3.50 cm. Set up a table to record the position of the lens x con and the po si tion of the screen (or im age) x im, both in cm. For the first two points, you fix the screen po si tion and find lens po si tions that gives fo cused im ages. For the other points, you fix the lens po si tion and move the screen to focus. 1. Place the screen at the end of the optical bench opposite the light, lightly tighten the screw, and record the position to the

6 nearest hun dredth of a centimeter (You have to read a vernier scale.). 2. Plug in the lighted object. 3. Place the lens as close as possible to the screen. Then move the lens away from the screen until a sharp image (small and bright) of the arrows appears on the glass screen (Fig. 3). Observe the image from the back of the screen while moving the lens, not from the same side as the lens. Figure 3 Fo cus ing a con verg ing lens near the screen (im age). 4. Lightly tighten the screw on the lens fixture and record the position of the lens to pair with the position of the screen. 5. Put the lens adjacent to the object. Move the lens away from the bulb until a sharp image (large and faint) appears on the screen (Fig. 4). (This may be a two-person job.) Tighten the screw and record the lens position, as well as the image position, again. Figure 4 Fo cus ing a con verg ing lens near the light bulb (ob ject). 6. Set the lens position at cm. Adjust the screen to get the best image. Record the screen position. This is your third pair of data values.

7 7. Set the lens at the following positions x con, and find the cor re spond ing fo cused x im : 21.00, 22.00, 23.00, 24.00, 25.00, 26.00, 27.00, 28.00, 29.00, 30.00, 35.00, 40.00, 45.00, 50.00, 60.00, 70.00, and Converging lens: Image of infinite objects 8. For the second technique of finding the focal length, put your lens and bench fixture on the optical bench that has a screen but no mounted lenses. If it is not obvious where it is, ask your TA. a. Record the screen position. b. Focus on a far away object through the window. Record the lens position. c. Move the lens and refocus. Write down the new lens position. d. Repeat step b until you have at least ten lens positions. Each partner should take at least five readings. Diverging lens: Images of many objects The object lamp on your original bench should not have moved. Set up a data table to record the position of the diverging lens x div and the po si tion of the screen (or im age) x im, in cm. Part 1 9. Set the diverging lens at cm. 10. Put the converging lens fixture as close as possible to the diverging lens and record the lens positions x div and x con. The pos i tive dif fer ence be tween the two po si tions is D. For the rest of your mea sure ments, the lenses are moved as one unit and D is con stant, so re cord ing x div im plic itly gives you x con. 11. Adjust the screen to get a sharp image and record its position x im. 12. Slide the diverging lens fixture out to cm, keeping the converging lens fixture abutted against its right side. Find the focused image, and record the positions in your data table. 13. Find image positions for x div = 18.00, 19.00, 20.00, 22.50, 25.00, 30.00, 35.00, 40.00, 50.00, and cm. Part Now put the converging lens to the left of the diverging lens instead of the right. Move the diverging lens to cm and record this x div. Mea sure and re cord x con with the fix tures snug against each other. D should not change ap pre cia bly but it does n t hurt to check. Find x im that gives a fo cused image. 15. Find image positions for x div = 41.00, 42.00, 43.00, 44.00, 45.00, 50.00, 55.00, 60.00, 70.00, and cm. Part Remove the two lenses from their fixtures. Look through both lenses at a distant object, then at something on the lab bench (if not the lab bench itself). Focus by moving the farther lens. 17. Which lens would the nearer lens have to be so that the combination acts as a microscope? As a telescope?

8 Analysis Converging lens 1 Start the Generic Lab Spreadsheet and enter the converging lens data into appropriately labeled columns. Calculate d o = x con -x obj, d i = x im - x con, 1/d o, and 1/d i in adjacent columns. 2. Plot 1/d i vs. 1/d o and in clude a trendline. 3. Use LINEST to find the y-in ter cept 1/f i and its er ror. 4. To find the x-intercept and its error, perform a LINEST cal cu la tion again, switch ing the x and y val ues. This cal cu lated y-in ter cept is 1/f o. 5. Find the average of the two intercepts, b=(1/f i +1/f o )/2. a. Propagate the error of this and calculate it (call it u{b 1 }). b. Find the standard error of the two intercepts and call it u{b 2 }. c. Your experimental value of the focal length is the inverse of b. Find this f. d. The error associated with this focal length is f times the larger of u{b 1 } and u{b 2 }, di vided by b. e. Compare the experimental value of the focal length with the nominal value of 13.0 ± 0.5 cm. 6. Find the average of the lens positions measured in step 8 of the Procedure. a. Find the uncertainty of this value. b. Compare this determination of the focal length with the value obtained in step 5. Diverging lens Part 1 D = x con - x div for the first point. d o div = x div - x obj d i conv = x im - (x div + D) Find d o con us ing the lens equa tion. d i div = D - d o con Part 2 D = x div - x con for the first point. d o con = x div - D - x obj d i div = x im - x div Find d i con us ing the lens equa tion. d o div = D - d i con Note that in steps 11 and 12 below, you need to use the values of 1/d o div and 1/d i div from both Part 1 and Part 2. That is, you are an a lyz ing one com bined data set, not two sep a rate ones. 7. In the same spreadsheet, enter the object position and the initial lens positions from Steps 9 & 10 in the Procedure and calculate D. 8. Enter your data of the diverging lens positions x div and the screen po si tions x im in la beled col umns for both Parts 1 and In adjacent columns, calculate d o div, d i conv, d o con (one of these two from the lens equa tion), and d i div us ing your data. 10. In adjacent columns, calculate 1/d o div and 1/d i div. 11. Plot 1/d i div vs. 1/d o div and in clude a trendline.

9 12. Use LINEST to find 1/f i and 1/f o for the di verg ing lens. 13. Repeat step 5 of the Analysis to find the diverging lens focal length and its uncertainty. Compare to ± 0.5 cm. Questions 1. For a converging lens, where must the object be placed to give a real image of the same size as the object? 2. For a converging lens, where must the object be placed to give an image at infinity? General Physics 3. Let D rep re sent the dis tance be tween an ob ject and its real im age formed by a con verg ing lens of fo cal length f. Show that D = 4 f is the min i mum sep a ra tion which can oc cur be tween the ob ject and its im age. What are the ob ject and im age dis tances which produce this minimum separation?

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