Lens Equation

(last edited June 14, 2013)

Dr. Larry Bortner

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 interface surface between two media at a non-perpendicular angle, its direction changes. This bending of a light ray is called refraction.

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 differently-sized 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. Different sizes of the spheres defining the sections cause the light to spread 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 do between an object and a lens, the distance di between the lens and the image position, and the focal length f of the lens. (A quick aside: the variables denoting the object and image distances vary widely and what is used here may not be the same as your text. The ideas are the same.) This basic relationship can be expressed by the thin lens equation:

(1)    In this experiment, several pairs of object and corresponding image distances are measured to check the functional 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 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. Figure 1 Converging lens ray diagram where the object is greater than the focal length. Figure 2 Ray diagram for a converging lens where the object is closer than the focal point.

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 ratio of the image size or height to the object height. A negative magnification signifies an inverted image. By this convention, a positive height indicates the arrow is erect and a negative height means the arrow is pointing down (valid for both object and image). From the similar triangles with angle α shown in Figs. 1 and 2, we can relate the magnification to the object and image distances:

(2)    Using the triangles with angle β in Fig. 1, we can come up with yet another expression for the ratio of the heights:

(3)    Dividing the last equation by di gives

(4)    and solving for the inverse focal length gives Eq. 1, the lens equation.

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 do is positive), since

(5)    This means that the image formed by a diverging lens with a real object is always a virtual image. The magnification is given by Eq.(3).

Ray diagrams for a diverging lens can also 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/di vs. 1/do   or 1/do vs. 1/di. The y-intercept for both of these graphs is the negative reciprocal of the focal length. Image of Infinite Object 2.   Focus on a object far enough away that we can assume that 1/do is zero compared to 1/di. Then the image distance is the focal length.

Determining the focal length of a diverging lens is more complicated, since the lens 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 two lenses, the object, and the screen fall along a straight line. (Experimentally, they are on an optical bench.) Further 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 fcon are the positions xobj, xdiv, xcon, and xim of the object, the two lenses, and the image. First we work backwards to find the object position of the converging lens, then use this as the image position of the diverging lens.

 1.     We find the object distance of the diverging lens and the image distance of the converging lens: (6)  2.     Use the lens equation to find the object distance of the converging lens: (7)  3.     Let D = xcon − xdiv be the distance between the two lenses. The object of the converging lens is the image of the diverging lens. So (8)        For a single point calculation we could then use the lens equation to find the focal length fdiv. This is not the best way to do things experimentally. We use the Images of Many Objects method: Find a bunch of values of do div and di div and plot 1/do div vs 1/di div (or vice versa). The y-intercept is 1/fdiv.

Procedure

You need the following items: 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 xobj = 3.50 cm. Keep it there. The fixture is not set at exactly 3.50; the plastic slide that is the object is offset from the fixture reading. This has been accounted for in the fixture placement.

The glass plate, the plastic slide that projects the arrow’s image, and the lens should be perpendicular to the optical bench length.

Converging Lens: Images of Many Objects

1.     Note and record that the object is at 3.50 cm.

2.     Set up a table on your data sheet to record the position of the lens xcon and the position of the screen (or image) xim, both in cm.

For the first two points, you fix the screen position and find lens positions that gives focused images. For the other points, you fix the lens position and move the screen to focus.

3.     Place the screen at the end of the optical bench opposite the light, lightly tighten the screw, and record the position to the nearest hundredth of a centimeter (You have to read a vernier scale.).

4.     Plug in the lighted object.

5.     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 Focusing a converging lens near the screen.

 6.     Lightly tighten the screw on the lens fixture and record the position of the lens to pair with the position of the screen. 7.     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 Converging lens focused near the light bulb. Checkpoint! Have the TA check your measurements before you proceed further.

8.     Set the lens position at 20.00 cm. Adjust the screen to get the best image. Record the screen position. This is your third pair of data values.

9.     Set the lens at the following xcon positions, and find the corresponding focused xim : 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 80.00.

Converging Lens: Image of Infinite Objects

10.  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

11.  The object lamp on your original bench should not have moved. Set up a data table to record the position of the diverging lens xdiv and the position of the screen (or image) xim, in cm.

Part 1: Diverging Lens Closer to Lamp

12.  Set the diverging lens at 16.00 cm.

13.  Put the converging lens fixture as close as possible to the diverging lens and record the lens positions xdiv and xcon. The positive difference between the two positions is D. (You will calculate this value in the Analysis.) For the rest of your measurements, the lenses are moved as one unit and D is constant, so recording xdiv implicitly gives you xcon.

14.  Adjust the screen to get a sharp image and record its position xim .

15.  Slide the diverging lens fixture out to 17.00 cm, keeping the converging lens fixture abutted against its right side. Find the focused image, and record the positions in your data table.

16.  Find image positions for xdiv= 18.00, 19.00, 20.00, 22.50, 25.00, 30.00, 35.00, 40.00, 50.00, and 60.00 cm.

Part 2: Converging Lens Closer to Lamp

17.  Now put the converging lens to the left of the diverging lens instead of the right. Move the diverging lens to 40.00 cm and record this xdiv. Measure and record xcon with the fixtures snug against each other. D should not change appreciably but it doesn’t hurt to check. Find xim  that gives a focused image.

18.  Find image positions for xdiv = 41.00, 42.00, 43.00, 44.00, 45.00, 50.00, 55.00, 60.00, 70.00, and 80.00 cm.

Part 3

19.  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.

20.  Which lens would the nearer lens have to be so that the combination acts as a microscope? As a telescope?

Analysis

Converging Lens

1.     Click on Start> Templates> Third Quarter> Lens Equation. Enter the converging lens data into the appropriate columns. Calculate do = xcon − xobj , di = xim − xcon, 1/do, and 1/di in adjacent columns.

2.     Plot 1/di vs. 1/do and include a trendline. Do not include error bars.

3.     Use LINEST to find the y-intercept 1/fi  and its error.

4.     To find the x-intercept and its error, perform a LINEST calculation again, switching the x and y values. This calculated y-intercept is 1/fo.

5.     Find the average of the two intercepts and call it b.

a.     Propagate the error of this calculation and calculate it (call it u{b1}).

b.     Find the standard error of the two intercepts and call it u{b2}.

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{b1} and u{b2}, divided by b.

e.     Compare the experimental value

6.     Find the average of the lens positions measured in step 9 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. Checkpoint! Have the TA check your calculations before you proceed further.

Diverging Lens Part 1

D = xcon −  xdiv for the first point.

do div = xdiv − xobj

di conv = xim − (xdiv + D)

Find do con using the lens equation.

di div = D − do con

Diverging Lens Part 2

D = xdiv  − xcon for the first point.

do con = xdiv − D − xobj

di div = xim − xdiv

Find di con using the lens equation.

do div = D − di con

Note that in steps 11 and 12 below, you need to use the values of 1/do div and 1/di div from both Part 1 and Part 2. That is, you are analyzing one combined data set, not two separate ones.

7.     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 xdiv and the screen positions xim for both Parts 1 and 2.

9.     For Part 1, using your data, calculate in this sequence: do div, di con, do con, and di div. Checkpoint! Have the TA check your calculations before you proceed further.

a.     For Part 2, using your data, calculate in this sequence: do con, di div, di con, and do div.

10.  In adjacent columns, calculate 1/do div and 1/di div. (The cell references are different for Parts 1 and 2.)

11.  Plot 1/di div vs. 1/do div and include a trendline. Do not include error bars.

12.  Use LINEST to find 1/fi and 1/fo for the diverging lens.

13.  Repeat step 5 of the Analysis to find the diverging lens focal length and its uncertainty. Compare to -20.0 ± 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?

Algebra-Based Physics

2.     For a converging lens, where must the object be placed to give an image at infinity?

Calculus-Based Physics

2.     Let D represent the distance between an object and its real image formed by a converging lens of focal length f. Show that D = 4×f  is the minimum separation which can occur between the object and its image. What are the object and image distances which produce this minimum separation?