phy lap

Lenses

NAME:

Date of Experiment : November 21st, 2014

Instructor: Salameh

1

Introduction

In this experiment, the images formed by different types of lenses will be classified and their focal lengths will be measured using the “quick and dirty method”, the lens equation, and the exact magnification method.

Images formed by lenses can be described by their magnification (larger, smaller, or same size), whether they are real or virtual, and whether they are upright, inverted or reverted (flipped on the horizontal axis). The lens equation shows the relationship between 3 different variables, the object distance (So), the image distance (Si), and the focal length, f. The lens equation shows

that !!" + ! !" =

! !. The object distance is the distance between the object in view and the

lens. The image distance is the distance between the lens and the image formed. If the object is far away from the lens, the distance value used in the equation is infinity. Therefore, if the value for So is replaced with infinity, the equation can be manipulated to

show that !! + ! !" =

! ! →

! !" =

! ! → Si = !. This manipulation of the lens equation describes

the “quick and dirty” method. This method enables an approximate calculation of the focal length by using the lens to form an image of object that is fairly far away. Then, the focal length is found to be equal to the image distance. However, the quick and dirty method can only be employed when using a converging lens. The method will not work for a diverging lens, as the image produced will be virtual and located between the object and lens. The equation for the magnification, M, of the image is related to both the image

distance (hi) and object distance (ho) through the equation M = !! !!. An image that is

inverted/upside down has a (-) sign; therefore a negative M value shows that the image is inverted. An image that is upright has a (+) sign; therefore a positive M value shows that the image remained upright. In addition, the height of the image and the distance of the image are proportional, as are the height of the object and the distance of the object.

Therefore, the magnification equation can be manipulated to show that M = − !"!". M does not have any units, therefore, it is usually written as Mx. The x denotes the degree of magnification. In addition, if an image has a magnification of exactly 1x, the focal length of the lens can be found by the equation Si + So = 4!. This equation applies to a single- lens system, where there is only one lens. In a two-lens system, the image from the first lens is the object from the second lens. In this manner, the focal length of a diverging lens can be found. The diverging lens must be placed with a converging lens that has a known focal length to calculate the focal length of the diverging lens.

2

Procedure Three lenses (A, B, and C) were placed on a labeled sheet of paper. The lens holders, the lamp, and the screen were assembled on the track. Lens A was placed on the track near the middle and the screen was placed at the other end. The track was placed on one of the group member’s shoulders aimed toward a far away light. The position of the lens was adjusted so that the image could be focused. The focal length was measured and recorded. The lens was then flipped and no change was observed. Next, this procedure was repeated for lens B. Lens B was placed in the lens holder and once again the focal length was determined. The lens system was then placed on the table and the object lamp was placed at one end. Lens A was once again placed inside the lens holder. Next, the lens and the screen were moved in order to focus the image. The object distance, object height, image distance, and image height were measured and recorded. This process was repeated for a total of 3 combinations of object distances and image distances for lens A. The images produced were recorded with a description. Next, lens B was placed in the lens holder. Next, the lens and the screen were moved in order to focus the image. The object distance, object height, image distance, and image height were measured and recorded. This process was repeated for a total of 3 combinations of object distances and image distances for lens B. The images produced were recorded with a description. Using lens A again, the object and image distanced were measured until the image was the same size as the object (with a 1x magnification). The object distance, image distance, object height, and image height were measured and recorded. This process was repeated for lens B, and the object and image distanced were measured until the image was the same size as the object (with a 1x magnification). The object distance, image distance, object height, and image height were measured and recorded. The images produced were recorded with a description. Next, the items on the track were rearranged and placed in the order of object lamp, lens C, lens A, and the screen. Lens C was placed before lens A as it was concave and thinner, representing a diverging lens. A small object distance was measured so that the image produced was in focus and clear. The image distances, object distances, image height and object height were measured from the point of the object lamp to both lens C and lens A. The images produced were recorded with a description.

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Data

Method Lens Ho (cm) Hi

(cm) So

(cm) Si

(cm) Focal Length

(cm) Magnification

M= hi/ho Image

Description Lens

Equation A 4

cm 10.5 cm

17 cm

38.5 cm 11.79 cm

M= hi/ho=10.5/4=2.62x

Larger,Inverted, Reverted

A 4 cm 2.3 cm

31.5 cm

17.8 cm 11.37 cm

0.575x Smaller,Inverted, Reverted

A 4 cm 1.0 cm

57.7 cm

15.4 cm 12.16 cm

0.25x Smaller,Inverted, Reverted

B 4 cm 1.8 cm

70.5 cm

30.5 cm 21.29 cm

0.45x Smaller,Inverted, Reverted

B 4 cm 10.2 cm

31 cm

72 cm 21.67 cm

2.55x Larger, Inverted, Reverted

B 4 cm 12.5 cm

29.6 cm

82 cm 21. 74 cm

3.13x Larger, Inverted, Reverted

Exact Magnification

A 4 cm 4

cm 23.5 cm

22.8 cm

So+Si=4f =(23.5+22.8)/4 =f=11.58 cm

1x Same size, Inverted,Reverted

B 4 cm 4

cm 41.7 cm

42.1 cm 20.95 cm

1x Same size, Inverted,Reverted

Diverging Lens in Tandem

A 4 cm 12.3 cm

9.7 cm

54.8 cm 8.24 cm

3.08x Larger, Inverted, Reverted

C 4 cm 12.3 cm

7.8 cm

56.7 cm 6.86 cm

3.08x Larger, Inverted, Reverted

Method Lens Focal Length (cm)

Quick and Dirty A 12 cm Quick and Dirty B 20 cm

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Calculations and Data Analysis Questions (2)

(2a)The average value of the focal length of lens A is calculated to be (11.79 cm + 11.37 cm + 12.16 cm)/3 = 11.77 cm. In addition, the graph of the reciprocals of the object and image distances (Si -1vs. S0 -1) gives an intercept, which equals 1/f. The intercept, 0.0836, gives the value of f as 1/0.0836 = 11.96 cm. (2b) The value of f determined from the graph (11.96 cm) is close to the average value calculated from the lens equation (11.77 cm). The discrepancy between the values can be attributed to the manner in which the values were obtained. The focal length value from the graph was obtained from the line of best fit, versus the average focal length value was obtained from a calculation. Since the uncertainty values were not calculated when manually calculating the average, the line of best fit already takes into account the uncertainty values. (2c) Since the average value for the focal length is a slightly skewed measure (due to the fact that the average is pulled higher or lower due to the largest and smallest values), the value obtained from the graph is a more accurate and reliable value for the focal length as it relies on the line of best fit (which takes into account the outliers and incorporates uncertainty). (3)

Lens Object Distance, So (cm)

Image Distance, Si (cm)

Focal Length (cm) ! !" +

! !" =

! !.

A 17 cm 38.5 cm 1/f = 1/17 + 1/38.5 = 0.085 f= 1/0.085 = 11.79 cm

A 31.5 cm 17.8 cm 1/f = 1/31.5 + 1/17.8 = 0.088 f = 1/0.088 = 11.37 cm

A 57.7 cm 15.4 cm 1/f = 1/57.7 + 1/15.4 = 0.082 f= 1/0.082 = 12.16 cm

Lens Object Distance, So (cm)

Image Distance, Si (cm)

Focal Length (cm) ! !" +

! !" =

! !.

B 70.5 cm 30.5 cm 1/f = 1/70.5 + 1/30.5 = 0.047 f= 1/0.0469 = 21.29 cm

B 31 cm 72 cm 1/f = 1/31 + 1/72 = 0.046 f = 1/ 0.046 = 21.67 cm

B 29.6 cm 82 cm 1/f = 1/29.6 + 1/82 = 0.045 f= 1/ 0.045 = 21.74 cm

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(3 contd.) The average value of the focal length of lens B is calculated to be (21.39 cm + 21.67 cm + 21.75 cm)/3 = 21.5 cm. In addition, the graph of the reciprocals of the object and image distances (Si -1vs. S0 -1) gives an intercept, which equals 1/f. The intercept, 0.0477, gives the value of f as 1/0.0477 = 20.96 cm. (4a) The value of f determined from the graph (20.96 cm) is not very close to the average value calculated from the lens equation (21.5 cm). However, the graph is a more reliable method of obtaining the value of the focal length that is most likely closest to the real focal length and the line of best fit can provide a better prediction of the 1/f value as opposed to the calculated 1/f value. (7a). While the graphing method is the most reliable method to obtain an accurate production of the focal length, the lens equation is the easier method as there is no graphing involved. The calculation to obtain the focal length is simple. The reciprocal of the focal length is obtained simply by plugging in the object and image distances into the formula !!" +

! !" =

! ! and the focal length is obtained by taking 1/(value for 1/f). One

source of error in the experiment that would have affected the manual calculations of the focal length to a greater extent is if the measurements of object distance and image distance were not taken from the center of the object lamp or apparatus on the track. Since the line of best fit takes into account some degree of uncertainty, these inaccurate values would have given an incorrect value for the calculated focal length and magnification. Another source of error stems from using an image that was not exactly focused. The measurements and calculations of object distance, image distance, image height, object height, focal length, and magnification rely on the image being focused to obtain the most accurate values. Using an unfocused image would have given incorrect values and therefore, an incorrect focal length. The value would have reflected the focal length and magnification of the unfocused image, not the focused image that was expected.

y"="$0.9617x"+"0.0836"

0"

0.01"

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0.08"

0" 0.01" 0.02" 0.03" 0.04" 0.05" 0.06" 0.07"

1/ Si %(1

/c m )%

1/So%(1/cm)%

1/Si%vs.%1/So%for%Lens%A%

y"="$1.0485x"+"0.0477"

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0.005"

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0.035"

0" 0.005" 0.01" 0.015" 0.02" 0.025" 0.03" 0.035" 0.04"

1/ Si %(1

/c m )%

1/So%(1/cm)%

1/Si%vs.%1/So%for%Lens%B%