If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
021. `Query 19
Question:
`qPrinciples of Physics and General College Physics Problem
24.54: What is Brewster's angle for an
air-glass interface (n = 1.52 for glass)?
Your solution:
Confidence rating:
Given Solution:
`aBrewster's angle is the smallest angle theta_p of
incidence at which light is completely polarized. This occurs when tan(theta_p) = n2 / n1,
where n2 is the index of refraction on the 'other side' of the interface.
For an air-glass interface, n1 = 1 so tan( theta_p) = n2 / 1
= n2, the index of refraction of the glass.
We get
tan(theta_p) = 1.52 so that
theta_p = arcTan(1.52).
This is calculated as the inverse tangent of 1.52, using the 2d
function-tan combination on a calculator.
WE obtain
theta_p = 56.7 degrees, approximately.
Self-critique (if necessary):
Self-critique Rating:
Question:
`qgen phy problem 24.44
foil separates one end of two stacked glass plates; 28 lines observed
for normal 670 nm light
gen phy what is the thickness of the foil?
Your solution:
Confidence rating:
Given Solution:
`aSTUDENT SOLUTION:
To solve this problem, I refer to fig. 24-31 in the text book as the
problem stated. To determine the thickness of the foil, I
considered the foil to be an air gap. I
am not sure that this is correct.
Therefore, I used the equation 2t=m'lambda, m=(0,1,2,...). THis is where the dark bands occur .
lambda is given in the problem as 670nm and m=27, because
between 28 dark lines, there are 27 intervals.
Solve for t(thickness):
t=1/2(27)(670nm)
=9.05 *10^3nm=9.05 um
INSTRUCTOR RESPONSE WITH DIRECT-REASONING SOLUTION:** Your
solution looks good. Direct
reasoning:
** each half-wavelength of separation causes a dark band so
there are 27 such intervals, therefore 27 half-wavelengths and the thickness is
27 * 1/2 * 670 nm = 9000 nm (approx) **
**** gen phy how many wavelengths comprise the
thickness of the foil?
GOOD STUDENT SOLUTION:
To calculate the number of wavelengths that comprise the thickness of
the foil, I use the same equation as above 2t=m'lambda and solve for m.
2(9.05 um)=m(6.70 *10^-7m)
Convert all units to meters.
m=27 wavelengths.
Self-critique (if necessary):
Self-critique Rating: