QUIZ 23,  CH.23  (REAL TEST 4)
 QUIZ 23  
SELECTED DISCUSSIONS TO exercises/problems BELOW.
YOU ARE RESPONSIBLE FOR ALL  PROBLEMS EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES. WE WILL REVIEW THEM IN-CLASS.
CH.  23
CH. 23  MULTIPLE CHOICE TBA  - Exercise/problems  2, 14, 15, 16, 19, 22, 24, 31, 42, 44, 49, 51, 52, 57, 68, 
TURN IN: 2, 14, 15, 22, 31, 42, 51, 52, 57, 68 
* DISCUSSIONS PROVIDED. 
2.   SEE LECTURE NOTES ; DISTANCE = c*t, where delay time t  is given. 
14. SEE FIGURE 23.2 AND MY LECTURE NOTES ; (a) E-vector in x direction means point  your right fingers in the x- direction and wrap E-vector into the y-axis, the direction of B-vector; clearly your thumb points along  positive z - axis, the propagation direction. Same method for parts b, c, d. See figures 23.5 AND 23.6. 
15. SEE LECTURE NOTES ; 

(a) FROM CH. 12 THEORY, WE KNOW THE PROPAGATION DIRECTION IS ALONG THE NEGATIVE X-AXIS: SEE EQUATION 23.5  AND SEE EQUATIONS 12.5 AND 23.3  FOR COMPARISON.  NOTE: EQUATION 23.3 CORRESPONDS TO EQUATION 12.5.

(b) c = f*lambda,  or f = c/lambda,  where f and lambda are the frequency and  wavelength. Note also f = w/(2*pi) where  w is the angular frequency shown in the expression for B . SEE EQUATION 23.5. 
f = c/lambda, where 2*pi/lambda = 1.38x10 4 rad/m  from page 766 discussion leading to equations 23.3. and 23.5.  NOTE: pi = 3.14...

(c) EQUATION  23.4 WILL GIVE  THE ELECTRIC FIELD AMPLITUDE FOR A FUNCTION THAT WILL LOOK SIMILAR TO THAT GIVEN FOR  B- FIELD. 
22. THE WAVES CARRY ENERGY AS SHOWN IN SECTION 23.5. SEE LECTURE NOTES ; SEE EQUATION 23.11  FOR THE INTENSITY I  (IN WATTS/m2) . ENERGY = I*AREA*TIME. The area and time are given. See equation 23.11 for I; note: Erms = Emax/(1.41..), where 1.41... = square root of 2. Note also that Erms = 0.707*Emax . 
31. n = c/v, where v is light's speed in material and c is the speed of light in a vacuum. v is easy : distance /time, both given.   
42. use geometry: (1)*sin theta 1 = (1.33)*sin theta 2; note:  tan theta 1  =  2.00/1.75 and  tan theta 2 = x/2.5,  where x is an unknown you are to find.  
51. NOTE: SOME BOOKS MAY SHOW A DIFFERENT PICTURE THAN THAT DESCRIBED IN THE PROBLEM, WITH DIFFERENT NUMBERS FOR EXAMPLE. Continue to use whatever numbers are  in your book and I will adjust my solution to either case.

 My book showed a cladding with n = 1.44 and a core with index 1.46. (Some books showed  1.46 and 1.48, respectively.) Internally, sin thetac =  1.44/1.46, if your book showed these numbers.  Solve for theta c. Next relate that angle to the initial angle theta shown in diagram:   Note, theta + theta c = 90 degrees. Thus:
sin  theta c  = cos theta.  Solve for theta, the largest angle allowed between the entering light ray and the longitudinal axis of fiber. As theta get larger, there is a reduction in the  angle between the  internal  ray and normal with top surface. If the latter angle get too small you will not have total internal  reflection.
52.  (a) (1)*sin 57.0  = n*sin theta 2, where theta 2 = 38.1 or 36.1. Find n in each case. (b) speed = c/n in each case.
57. Note: I should have mentioned in class the situation for un-polarized light incident on a polarizer. In that case you have to take the average of Io*cos2theta. The average of cos2theta = 1/2, thus the emerging light has intensity  Io/2 before it hits the second filter. After it passes through the second polarizer, the intensity is given by (Io/2)*cos241. What is the state of polarization after it passes through the second filter? In other words, along what axis will the light be finally polarized?
68. Force = (radiation pressure)* area, and radiation pressure  is given by equation 23.13. Use Newton's Second Law: acceleration = force/mass to determine if the person  experiences a discernable change in motion. Assume the person's mass is 70 kg.