#$&*
course Phy 202
12/18 3
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?
The radius of the cylinder is 5.00cm. Distance where central ray strikes= (1/4)(5cm)= 1.25cm. I calculated angle of incidence to be 14 degrees.
2. For the index of refraction you obtained, what therefore will be the angle of refraction for this ray?
Index of refraction of air= 1. Index of refraction of water= 1.33.
Snell’s Law: n1sin(theta1)=n2sin(theta2)
Sin(angle of refraction)= sin(14)/0.77
Angle of refraction= 18.3 degrees
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?
Difference between theta_i and theta_r = 4.3 degrees. Sin(4.3)= 0.075
Length of hypotenuse of the right triangle= ¼*(5cm)/0.075= 16.7cm
4. How far from the 'front' of the lens did the sharpest image form?
The sharpest image formed at 3.8cm from the lens.
5. Should the answer to #3 be greater than, equal to or less than the answer to #4 and why?
I would have thought the two numbers would be equal because both questions were asking to find the focal point.
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?
The refracted ray is 1.25cm away from the central ray when it strikes the back of the lens. The angle of incidence is 14 degrees and the angle of refraction is 18.3 degrees.
7. At what angle with the central ray does the refracted ray therefore emerge from the 'back' of the lens?
18.3 degrees
8. How far from the 'back' of the lens will the refracted ray therefore be when it crosses the central ray?
16.7cm from the back of the lens
"
&#This looks good. Let me know if you have any questions. &#