Experiment 12: Focal Length and Magnification of a Thin Lens Procedure: ???? Pla
ID: 2033295 • Letter: E
Question
Experiment 12: Focal Length and Magnification of a Thin Lens
Procedure:
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Place the light source and the screen on the optics bench 1 m apart with the light source’s crossed-arrow object toward the screen. Place the lens between them (see Figure 12.1).
Starting with the lens close to the screen, slide the lens away from the screen to a position where a clear image of the crossed-arrow object is formed on the screen. Measure the image distance and the object distance. Record these measurements (and all measurements from the following steps) in Table 12.1.
Measure the object size and the image size for this position of the lens.
Without moving the screen or the light source, move the lens to a second position
where the image is in focus. Measure the image distance and the object distance.
Measure the object size and image size for this position also. Note that you will not see the entire crossed-arrow pattern. Instead, measure the image and object sizes as the distance between two index marks on the pattern (see Figure 12.2 for example).
Repeat steps 2 and 4 with light source-to-screen distances of 90 cm, 80 cm, 70 cm, 60 cm, and 50 cm. For each light source-to-screen distance, find two lens positions where clear images are formed. (You don’t need to measure image and object sizes.).
Data:
Distance
do
di
1/do
1/di
Image size
Object size
Calculated Image Distance
Measured Magnification
% difference
100
87.9
12.1
0.0113
0.0826
0.297
2
0.137656428
0.1485
7.877272727
12
88
0.0833
0.0114
15.79
2
7.333333333
7.895
7.659090909
90
77.7
12.3
0.0129
0.0813
0.334
2
0.158301158
0.167
5.495121951
12.2
77.8
0.082
0.0129
12.949
2
6.37704918
6.4745
1.5281491
80
67.5
12.5
0.0148
0.08
0.39
2
0.185185185
0.195
5.3
12.5
67.5
0.08
0.0148
11.035
2
5.4
5.5175
2.175925926
70
57
13
0.175
0.0769
0.482
2
0.228070175
0.241
5.669230769
12.9
57.1
0.0775
0.0175
8.956
2
4.426356589
4.478
1.166725044
60
46.2
13.8
0.0216
0.0725
0.636
2
0.298701299
0.318
6.460869565
13.6
46.4
0.0735
0.02155
6.911
2
3.411764706
3.4555
1.281896552
50
34.9
15.1
0.0287
0.0662
0.946
2
0.432664756
0.473
9.322516556
15
35
0.0666
0.0286
4.844
2
2.333333333
2.422
3.8
*The image of the object was inverted*
Question:
Is the image real or virtual? How do you know and explain why, for a given screen-to-object distance, there are two lens positions where a clear image forms.
Distance
do
di
1/do
1/di
Image size
Object size
Calculated Image Distance
Measured Magnification
% difference
100
87.9
12.1
0.0113
0.0826
0.297
2
0.137656428
0.1485
7.877272727
12
88
0.0833
0.0114
15.79
2
7.333333333
7.895
7.659090909
90
77.7
12.3
0.0129
0.0813
0.334
2
0.158301158
0.167
5.495121951
12.2
77.8
0.082
0.0129
12.949
2
6.37704918
6.4745
1.5281491
80
67.5
12.5
0.0148
0.08
0.39
2
0.185185185
0.195
5.3
12.5
67.5
0.08
0.0148
11.035
2
5.4
5.5175
2.175925926
70
57
13
0.175
0.0769
0.482
2
0.228070175
0.241
5.669230769
12.9
57.1
0.0775
0.0175
8.956
2
4.426356589
4.478
1.166725044
60
46.2
13.8
0.0216
0.0725
0.636
2
0.298701299
0.318
6.460869565
13.6
46.4
0.0735
0.02155
6.911
2
3.411764706
3.4555
1.281896552
50
34.9
15.1
0.0287
0.0662
0.946
2
0.432664756
0.473
9.322516556
15
35
0.0666
0.0286
4.844
2
2.333333333
2.422
3.8
Explanation / Answer
The image is real. 1) because it is inverted. 2) As per your discription it is the experiment of conjugate focal method. Where screen and source are at fixed distance, image will form for two different positions of lens. In both the cases the object distance is more than focal length.
In this case image is real.
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