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course Phy 232
This experiment uses a cylindrical container and two lamps or other compact light sources. Fill a cylindrical container with water. The cylindrical section of a soft-drink bottle will suffice. The larger the bottle the better (e.g., a 2-liter bottle is preferable to a 20-oz bottle) but any size will suffice.
Position two lamps with bare bulbs (i.e., without the lampshades) about a foot apart and 10 feet or more from the container, with the container at the same height as the lamps. The line separating the two bulbs should be perpendicular to the line from one of the bulbs to the cylindrical container. The room should not be brightly lit by anything other than the two bulbs (e.g., don't do this in front of a picture window on a bright day).
The direction of the light from the bulbs changes as it passes into, then out of, the container in such a way that at a certain distance behind the container the light focuses. When the light focuses the images of the two bulbs will appear on a vertical screen behind the cylinder as distinct vertical lines. At the focal point the images will be sharpest and most distinct.
Using a book, a CD case or any flat container measure the distance behind the cylinder at which the sharpest image forms. Measure also the radius of the cylinder.
As explained in Index of Refraction using a Liquid and also in Class Notes #18, find the index of refraction of water.
Then using a ray-tracing analysis, as describe in Class Notes, answer the following:
1. If a ray of light parallel to the central ray strikes the cylinder at a distance equal to 1/4 of the cylinder's radius then what is its angle of incidence on the cylinder?
R of Cyl=4.9 cm
Distance=(1/4)*(4.9)= 1.225 cm
We can find the angle of incidence with these values.
sin(theta)=1.225cm/4.9cm
theta=sin^-1(.25)=14.5 deg
2. For the index of refraction you obtained, what therefore will be the angle of refraction for this ray?
Index of refraction of water,air: 1.33,1.0003
n1sin(theta1)=n2sin(theta2)
(1.0003)sin(14.5)=(1.33)sin(theta2)
theta2=10.85 deg
3. If this refracted ray continued far enough along a straight-line path then how far from the 'front' of the lens would it be when it crossed the central ray?
tan(14.5 degrees)=1.225cm/distance
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The incoming ray makes angle 14.5 degrees with normal, but the refracted ray makes angle 10.85 degrees with normal.
Neither of these is the angle of the refracted ray with central axis, which is the same as the angle of the refracted ray with the incoming beam (which after all is parallel to the central axis). So neither of these angles is directly helpful in determining where the refracted ray would, if it continued along a straight line, intersect the central ray.
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dist=4.74cm
4. How far from the 'front' of the lens did the sharpest image form?
4.5cm
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The cylinder has a diameter of 10 cm. 4.5 cm from the front of the lens wouldn't be halfway through the lens, and the image didn't form inside the cylinder.
You might mean 4.5 cm from the back of the lens, but that's not what the question asked.
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5. Should the answer to #3 be greater than, equal to or less than the answer to #4 and why?
These answers should be equilavent but are not exactly the same based on several errors.
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There is a second refraction before the actual ray crosses the central axis.
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6. How far is the actual refracted ray from the central ray when it strikes the 'back' of the lens? What is its angle of incidence at that point? What therefore is its angle of refraction?
dist-approx lens depth=4.74cm-.75cm=3.99cm
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The refracted ray approaches the central ray, so when it reaches the back of the lens it isn't further from the central ray than it was when it encountered the cylinder.
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Angle of inc: sin^-1(1.225cm/3.99cm)=17.9deg
Angle of ref: 13.4deg
7. At what angle with the central ray does the refracted ray therefore emerge from the 'back' of the lens?
17.9deg
8. How far from the 'back' of the lens will the refracted ray therefore be when it crosses the central ray?
The lens crosses the central ray at 3.99cm."
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You got the incident angle and the angle of refraction right.
After that you demonstrated good knowledge of trigonometry, but there were errors and oversights, so most of the subsequent analysis will need to be redone.
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