|QUIZ 23, CH.23 (REAL TEST 4)|
SELECTED DISCUSSIONS TO exercises/problems
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 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
(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.
(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.|