lab report

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Introduction
Convex lens and diverging lens are some of the types of thin lenses. The two affect the behaviour of a ray of light as it passes through them, either diverging the ray or beam that passes through them, in the case of a concave lens or converging the ray or beam of light as it passes through them at a certain point, in the case of a convex lens. The position of the object corresponding to the mirror affect the locus of the image made as rays of light emerge through the given thin lens.
Data
Exercise 1
do di do+di
28 72 100
30.5 59.5 90
40 40 80
29.6 65.3 94
28.7 72.1 100
Results
Exercise 1: Qualitative Investigation of Image Formation
When I envelop the top half of the converging lens the real image formed will remain unchanged, in position, and in alignment (upside down). A minimum of 2 light rays are required for the formation of an image in this case. With the top half of the convex lens wrapped, therefore, the light rays are obstructed in their way, the light beam moving from the tip of the object and parallel to the Principal Axis is hindered from passing through and from being refracted by the convex lens. The light ray moving from the highest point of the object through the main focal point of the converging lens is refracted and emerges from the lens moving parallel to the principal axis. The ray from the tip of the object travelling through the centre of curvature in a straight line remains unhindered as it travels through the convex lens, and is not refracted.

Its intersection with the other ray is sufficient for the formation of an image, therefore covering the top half of the converging does not affect the image.
Determining the focal length of a positive convex lens
(1 / f) = (1 / di) + (1 / do)
f = (di * do) / (di + do)
do = 28cm
di = 72cm
f = (28*72) / (100)
= (2016) / 100
= 20.16 cm
di = 59.5 cm
do = 30.5 cm
f = (30.5*59.5) / (59.5+30.5)
= (1814.75) / (90)
= 20.16 cm
di = 40 cm
do = 40 cm
f = (40*40) / (40+40)
= (80) / (80)
= 20 cm
di = 65.3 cm
do = 29.6 cm
f = (29.6 * 65.3) / (29.6 + 65.3)
= (1932.88) / (94.9)
= 20.56 cm
di = 72.1 cm
do = 28.7 cm
f = (28.7 * 72.1) / 100
= 20.69 cm
Mean focal length = (20.69 + 20 + 20.56 + 20.16 + 20.16) / 5
= 101.57 / 5
= 20.314 cm
Standard Deviation = √ (Variance)
Variance = (20.16-20.314)2 + (20.16 – 20.314) 2 + (20 – 20.314) 2 + (20.56 – 20.314) 2 + (20.69 – 20.314) 2
Variance = ((-0.154)2 + (-0.154)2 + (-0.314) 2 + (0.246) 2 + (0.376) 2) / 5
= 0.34792 / 5
= 0.069584
√ 0.069584 = 0.02638
Data
Exercise 2
x- x+ xs Diameter of image
2 28.2 100 9.5
12 34.2 100 5.5
27.3 52.9 100 2.1
10.8 36.8 90 4.1
20.3 45.25 95 2.5
Results
      Converging Lens Diverging Lens
X- X+ Xs di do f di do f
2.0000 28.2000 100.0000 71.8000 27.7220 20.0000 -1.5220 2.0000 -6.3683
12.0000 34.2000 100.0000 65.8000 28.7336 20.0000 -6.5336 12.0000 -14.3429
27.3000 52.9000 100.0000 47.1000 34.7601 20.0000 -9.1601 27.3000 -13.7858
10.8000 36.8000 90.0000 53.2000 32.0482 20.0000 -6.0482 10.8000 -13.7465
20.3000 45.2500 95.0000 49.7500 33.4454 20.0000 -8.4954 20.3000 -14.6092
Exercise 2: Calculating the Focal Length of a Negative Lens
With the concave lens, light source and the screen provided I cannot produce an image, which can be fixated on display, since images formed by a concave lens are virtual (formed beside the object), hence making it difficult to be fixated on the screen.
Mean Focal Length = (6.3683 + 14.3429 + 13.7858 + 13.7465 + 14.6092) / 5
= 62.8526 / 5
= 12.5705
= -12.5705 cm
Variance = ((6.3683 – 12.5705)2 + (14.3429 – 12.5705)2 + (13.7858 – 12.5705)2 + (13.7465 – 12.5705)2 + (14.6092 – 12.5705)2 ) 5
= ((-6.2022)2 + (1.7724)2 + (1.2153)2 + (1.176)2 + (2.0387)2 ) / 5
= (48.62491) / 5
= 9.724983
√ 9.724983 = 3.1185
Exercise 3: A Ray Diagram
The final image position is at 89.5 cm mark. Using the ray diagram scale, for the diverging lens, di = 1.7 cm, do = 4 cm. and for the converging lens di = 8.75 cm, do = 6.8 cm. These values translate to, for the diverging lens, di = 8.5 cm, do = 20 cm. and for the converging lens di = 43.75 cm, do = 34 cm. In this experiment the image position is at 89.5 cm compared to 95 cm according to the experimental values.
Exercise 4: Magnification
Case 1: Configuration as used in exercise 3
m = (yi / yo)
= 2.5 / 4
= 0.625
Case 2: Value calculated using thin lens equation
m = (di-/ do-) * (di+ / do+)
= -(49.75 cm / 33.4454 cm) * -(8.4954 cm / 20.3 cm)
= -1.487499 * -0.41849
= 1.487499 * 0.41849
= 0.6225
Case 3: Values as obtained from ray diagram
m = (di-/ do-) * (di+ / do+)
= (1.7 cm / 4 cm) * (8.75 cm / 6.8 cm)
= 0.425 * 1.2868
= 0.54689
(round off)
= 0.5469
Comparing case 1 and case 2 respectively, a magnification of 0.625 is more than 0.6225 by 0.0025, Comparing case 1 and case 3, in case 1 the value of magnification is greater than that of case 3 by 0.00756, Comparing case 2 and case 3, the value of magnification for case 2 is greater than of case 3 by 0.07561.
Conclusion
For a diverging lens, the image created is usually upright, lesser than the object (amplification less than 1 but greater than 0) and virtual (difficult to focus on a screen). For a convex lens, when the object is further than the pivotal point the image is upside down and real (can be focused on screen). Using basic principles of image formation, one can locate the position of an image by drawing a ray diagram. A minimum of two rays is sufficient in locating the position of an image when drawing a ray diagram.

Works Cited
Thin lenses https://phys.libretexts.org/TextMaps/General_Physics_TextMaps/Map%3A_University_Physics_(OpenStax)/Map%3A_University_Physics_III_(OpenStax)/2%3A_Geometric_Optics_and_Image_Formation/2.4%3A_Thin_LensesThin Lens Equation www.hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenseq.html
Thin Lenses www.physics.louisville.edu/cldavis/phys299/notes/lo_lenses.html