q_q_assignment19

course Phy 202

ƒµùå·˜§”Á«£W×íÞÎê´¿ý^š¨Ü¿}assignment #019

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

ñCèËêBy×ßïçb¾„˜‚´zþ˜ž”ª

Physics II

07-25-2006

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01:06:37

Principles of Physics and General College Physics Problem 24.54: What is Brewster's angle for an air-glass interface (n = 1.52 for glass)?

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RESPONSE -->

tan( theta)=n

tan(theta)=1.52

56.66 degrees

.................................................

......!!!!!!!!...................................

01:07:42

Brewster'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.

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RESPONSE -->

i got the same answer

.................................................

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01:12:03

gen phy problem 24.43 foil separates one end of two stacked glass plates; 28 lines observed for normal 650 nm light

gen phy what is the thickness of the foil?

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RESPONSE -->

lambda=6.7*10^-7m

number of lines is 28 +1 on the end=29

2t=m*lambda

2t=(29)*(6.7*10^-7m)

t=1.943*10^-5/2=9.715*10^-6m

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01:13:53

STUDENT 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(2)(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) **

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RESPONSE -->

I misjudged the intervals and thought it was saying 28 + the end one. Once I changed that, I got a similar answer.

.................................................

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01:14:05

**** gen phy how many wavelengths comprise the thickness of the foil?

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RESPONSE -->

.................................................

......!!!!!!!!...................................

01:14:50

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.

......!!!!!!!!...................................

RESPONSE -->

My quary skipped the question and went directly to the answer. I do understand how it was found.

................................................."

Good job. Let me know if you have questions.

q_q_assignment19

course Phy 202

ƒµùå·˜§”Á«£W×íÞÎê´¿ý^š¨Ü¿}assignment #019

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

ñCèËêBy×ßïçb¾„˜‚´zþ˜ž”ª

Physics II

07-25-2006

......!!!!!!!!...................................

01:06:37

Principles of Physics and General College Physics Problem 24.54: What is Brewster's angle for an air-glass interface (n = 1.52 for glass)?

......!!!!!!!!...................................

RESPONSE -->

tan( theta)=n

tan(theta)=1.52

56.66 degrees

.................................................

......!!!!!!!!...................................

01:07:42

Brewster'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.

......!!!!!!!!...................................

RESPONSE -->

i got the same answer

.................................................

......!!!!!!!!...................................

01:12:03

gen phy problem 24.43 foil separates one end of two stacked glass plates; 28 lines observed for normal 650 nm light

gen phy what is the thickness of the foil?

......!!!!!!!!...................................

RESPONSE -->

lambda=6.7*10^-7m

number of lines is 28 +1 on the end=29

2t=m*lambda

2t=(29)*(6.7*10^-7m)

t=1.943*10^-5/2=9.715*10^-6m

.................................................

......!!!!!!!!...................................

01:13:53

STUDENT 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(2)(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) **

......!!!!!!!!...................................

RESPONSE -->

I misjudged the intervals and thought it was saying 28 + the end one. Once I changed that, I got a similar answer.

.................................................

......!!!!!!!!...................................

01:14:05

**** gen phy how many wavelengths comprise the thickness of the foil?

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

01:14:50

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.

......!!!!!!!!...................................

RESPONSE -->

My quary skipped the question and went directly to the answer. I do understand how it was found.

................................................."

Good job. Let me know if you have questions.