Lab10_VanLe
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Richland Community College *
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Date
Jan 9, 2024
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docx
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Uploaded by CountApePerson449
Van Thin Le
Professor Xiang Song
Phys-2426-51700
17 November 2023
Lab 10: Geometric Optics Part A Theory Please study concept of geometric optics to answer the following questions. 1.
What is the image, real image, and virtual image?
An image represents visual perception, like a photo or a two-dimensional picture, mimicking a physical object.
Real Image: Formed by the convergence of light rays, can be projected onto a screen.
Virtual Image: Appears to diverge from a point, cannot be projected, formed by the apparent path of light rays.
2.
Write and describe thin lens equation, lens make’s equation, and magnification equation. Give the meaning of positive or negative sign for each quantity.
Thin Lens Equation:
-
Relates object distance (p), image distance (q), and focal length (f) of a lens: 1
f
=
1
q
+
1
p
-
Positive/negative signs:
o
u and v are positive if measured from the side where light enters the lens.
o
f is positive for converging lenses (like convex) and negative for diverging lenses (like concave).
Lens Maker's Equation:
-
Links focal length (f), refractive index (n), radii of curvature (R1, R2), and lens thickness (t).
1
f
=(
n
−
1
)
(
1
R
1
−
1
R
2
)
-
Positive/negative signs:
o
R1 and R2 are positive if the center of curvature faces incoming light.
o
t is positive for a thicker lens center, negative for a thinner center.
Magnification Equation:
-
Calculates image-to-object size ratio. m
=
h
'
h
=
−
v
u
-
Positive magnification = upright image; negative = inverted.
-
v positive for real images, negative for virtual; u positive for objects where light comes from, negative otherwise.
3.
A converging lens has magnitude radius of curvature R = 78 cm for both surfaces and refractive index n = 1.60. An object is placed 35.0 cm from the lens. Find focal length f, the image distance q, magnification M, and describe the image. 1
f
=(
n
−
1
)
(
1
R
1
−
1
R
2
)
¿
(
1.6
−
1
)
(
1
0.78
−
−
1
0.78
)
=
1.53
→f
=
0.65
m
1
f
=
1
q
+
1
p
→
1
q
=
1
f
−
1
p
=
1
0.65
−
1
0.35
=
1.318
m
→q
=−
0.758
m
m
=
−
q
p
=
−
0.758
0.35
=
2.16
The focal length is 65 cm, the Image distance is -75.8 cm, the magnification is 2.16 , and the
image is virtual and inverted.
4.
Two thin lenses of one converging lens f
1
= 30.0 cm and one diverging lens f
2
= −
20.0 cm are separated by d
= 10.0 cm as illustrated in the figure. An object is placed 45.0 cm to the left of lens Find the position and the magnification of the final image.
q
1
=
f
1
p
1
p
1
−
f
1
=
30
×
45
45
−
30
=
90
cm
m
1
=
−
q
1
p
1
=
−
90
45
=−
2
p
2
=
d
−
q
1
=
10
−
90
=−
80
c m
q
2
=
f
2
p
2
p
2
−
f
2
=
(−
20
)
×
(−
80
)
−
80
−(−
20
)
=−
26.7
cm
m
2
=
−
q
2
p
2
=
26.7
−
80
=−
0.33
M
=
m
1
m
2
=−
2
×
−
0.33
=
0.66
Part B Lab Go to PhET
website. Click on Simulation/Physics. Under Physics, choose Light & Radiation. Under Light & Radiation, Geometric Optics
is the first simulation (Location might change. A to
Z search.). Click to run the Geometric Optics
. <1> Converging Lens Basic Operation
for
Geometric Optics/Lens Simulation: Choose converging lens (in default at the middle top of screen). Place check marks on Principal Rays, Focal Points, Virtual Image, and Labels. Keep default values for Curvature Radius R = 80 cm (R
1
= 80 cm and R
2
= −80 cm). Set Refractive Index of lens n = 1.5 and Diameter of Lens to max D = 120 cm. Place cursor on pencil (object on left side) and move up till eraser tough the Optical Axis (central line). Test and understand all functional tools on screen. You must practice figuring out what is the best way to complete the measurement. Data Table 1: Converging (Convex) Lens Using Lens Maker’s Equation 1/f = (n – 1) • (1/R
1
– 1/R
2
) to find focal length: f = 80cm
Where Refractive Index of lens n = 1.5, Curvature Radius R
1
= 80 cm and R
2
= −80 cm. Drag and arrange ruler and object to object position P = 200 cm (P at left side of lens). Drag ruler to measure image position Q (measured value of image position). Using P and f to calculate image position Q
C
: Q
C
=Pf/(P−f) Calculate magnification M: M =−Q/P Comment type of image: Real or Virtual, Enlarged or Reduced, Inverted or Upright Attach Screenshot of P = 200 cm Measurement from the Simulation to Lab Report
. P (cm) Q (cm) Q
C
=P•f /(P−f) (cm) M =−Q/P Image Type 200 133
133.33
-0.665 Real, Reduced, Inverted
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140 187 186.67
-1.336
Real, Enlarged, Inverted
44 -98
-97.78
2.23
Virtual, Reduced, Upright
Question: •
Show your work on calculation of focal length. If Curvature Radius of lens R = 80 cm, why are R
1
positive and R
2
negative? 1
f
=
(
n
−
1
)
(
1
R
1
−
1
R
2
)
¿
(
1.5
−
1
)
(
1
80
−
1
−
80
)
=
0.0125
→f
=
80
cm
Since r1 lies on right it is positive, and since r2 lies on left it is negative.
•
Give detail explanation the type of image for P = 44 cm. The Image distance is -98 cm, the magnification is 2.33 , and the image is Virtual, Reduced,
Upright
Data Table 2: Converging Lens, Varying Focal Length f Referring to Basic Operation, vary Refracted Index n (as listed in Table) and Curvature Radius R (R
1
= + R and R
2
= −R). Calculate focal length, f, using 1/f = (n – 1) • (1/R
1
– 1/R
2
). Set
Diameter as 120 cm. Place object at position P (as listed Table). Measure image position Q. Calculate magnification M =−Q/P.
Attach a Screenshots of First and Last Data from the Simulation to Lab Report
. n R (cm) f (cm) P (cm) Q (cm) M =−Q/P 1.53 80 75
200 122
-0.61
1.80 80 50
200 66
-0.33
1.63 83 66
110 164 -1.49
1.63 98 78
40 -83 2.08
Question: •
Show your work on calculation of focal length for first and last data rows. For the first data row:
1
f
=
(
1.53
−
1
)
(
1
80
−
1
−
80
)
=
75.5
cm
For the last data row:
1
f
=
(
1.63
−
1
)
(
1
98
−
1
−
98
)
=
77.8
cm
•
Comment (give reasons) properties of image for first and last data rows. -
For the first data row:
M = -0.61, so the image is real, inverted, and reduced. -
For the last data row:
M = 2.8, so the image is virtual, upright, and enlarged. <2> Diverging Lens Basic Operation
for
Geometric Optics/Lens Simulation: Choose the diverging lens symbol located at the middle top of the screen. Place check marks on Principal Rays, Focal Points, Virtual Image, and Labels. Keep default values for Curvature Radius R = 80 cm (R
1
= −80 cm and R
2
= 80 cm). Set Refractive Index of lens n = 1.5 and Diameter of Lens to max D = 120 cm. Place cursor on pencil (object on left side) and move up till eraser tough the Optical Axis (central
line). Test and understand all functional tools on screen. You must practice figuring out what is the best way to complete the measurement. Data Table 3: Diverging (Concave) Lens Using Lens Maker’s Equation 1/f = (n – 1) • (1/R
1
– 1/R
2
) to find focal length: f = -80cm
Where Refractive Index of lens n = 1.5, Curvature Radius R
1
= −80 cm and R
2
= 80 cm. Drag and arrange ruler and object to object position P = 200 cm (P at left side of lens). Drag ruler to measure image position Q. Using P and f to calculate image position Q
C
: Q
C
=Pf/(P−f) Calculate magnification M: M =−Q/P Comment type of image: Real or Virtual, Enlarged or Reduced, Inverted or Upright Attach Screenshot of P = 200 cm Measurement from the Simulation to Lab Report
.
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P (cm) Q (cm) Q
C
=P•f /(P−f) (cm) M =−Q/P Image Type 200 -58 -57.1 0.29
Virtual, Reduced, Upright 140 -51 -50.9
-0.36 Virtual, Reduced, Upright 44 -29
-28.4 -0.66
Virtual, Reduced, Upright Question: •
Show your work on calculation of focal length. If Curvature Radius of lens R = 80 cm, why are R
1
positive and R
2
negative? 1
f
=
(
n
−
1
)
(
1
R
1
−
1
R
2
)
¿
(
1.5
−
1
)
(
1
−
80
−
1
80
)
=−
0.0125
→f
=−
80
c m
•
Give detail explanation the type of image for P = 44 cm. The Image distance is -29 cm, the magnification is -0.66, and the image is Virtual, Reduced,
Upright
Part C Discovery Beyond what you have done with Geometric Optics
Simulation, make few new findings using different function and tools, or give real life examples for lens and image. You can express your findings by data table, word, equations, and/or screenshot graph/Video, but do giving physics explanation. You could also design meaningful experiment and show it here. In Part C minimum 10 sentences are required.
Lenses, like those in cameras and eyeglasses, manipulate light to create images. These representations, from photographs to microscope views, result from light interacting with lenses or mirrors. For instance, a camera lens focuses light onto sensors or film to capture scenes. Factors like aperture and focal length influence the resulting image. A magnifying glass, a convex lens, thickens at its center, converging light to enlarge objects. Conversely, convex lenses, thicker in the middle, converge light rays, crucial in magnifying glasses, cameras, and correcting vision issues. They form real images by converging refracted light at a focal point. These lenses enable applications like focusing light for maximum magnification and precise imaging, evident in microscopes. Their ability to concentrate light impacts various fields, including photography and vision correction.