Please answer all these questions and give the correct option with just a little explanation/calculations. I need the answers ASAP within an hour

QUESTION 10* This question and the next two refer to this situation: A monochromatic laser beam is polarized in the incident x- direction and propagates in the positive : polarization direction along the z-axis as shown in the figure. 450 It passes through an ideal quarter wave plate at the origin. The fast and slow axes are oriented at quarter wave plate 45" relative to the x-axis. Then, at = = d the light (fast axis shown) beam passes through an ideal linear polarizer with the polarization direction tilted at 45" 450 relative to the x-axis as shown in the figure. linear polarizer The intensity of the beam after it passes through (polarization the quarter wave plate is direction shown) O(a) equal to the incident intensity. O(b) one half the incident intensity. O(c) one quarter the incident intensity. Z QUESTION 11**# Assume the maximum amplitude of the electric field of the incident beam before passing through the quarter wave plate is En- After passing through the linear polarizer located at = = d, a possible expression for the electric field of the wave is O(a) D(b) a. B(z.() = - &cos (1z - ax)-9cos(12-20)] O(c) O(d) b. Ez!) = 20 [ Roos ()z + at ) + poos (iz tor)] D(e) C. BIZ!) = [&cos ( )z + ax ) +9cos (kz -20) ] d E(z,t) = Rcos (kz - or ) + 9sin (1z -20)] QUESTION.1 2aided by 100000843892771 from CourseHere.com on (4-26-2024 08.22:43 GMT -05:00 hips:/courses.physics.illinois.edu'phys212/5p2018/practice/practice.pllexama/sp10 49 http:/www.courschern.com/file/3048822 1/Spring-2010-Physics-212-Hour-Exam-3pdf? 4/262018 Spring 2010 Physics 212 Hour Exam 3 If Eo is 1000 V/m and the beam area is 10 me, what is the average power of the incident light beam? O(a) Pave = 0.33 mW O(b) Pave =0.57 mW O(c) Pave =0.67 mW O(d) Pave = 1.33 mW O(e) Pave = 1.67 mWQUESTION 8* This question and the next one concern the following situation: A circular capacitor is hooked up to some external electronics that cause a constant current / to flow into it for a brief period of time. This problem studies what happens in the vicinity of the capacitor during that time period. The figures below shows a perspective and a side view of the capacitor and the wires coming in and out of it. The radius of the capacitor is R, the gap between the plates is d and R > > d. This means you can assume that the electric field in the capacitor is uniform and directed straight across the gap, although it may change with time. Three points, labeled 1, 2 and 3, are located a distance R/2 from the center line, with points 1 and 3 outside and point 2 inside the capacitor. During the time interval that the constant current / is flowing through the capacitor, the magnetic field at point 2 is O(a) directed out of the page. perspective view O(b) directed into the page. R L R/2 side view QUESTION 9** Compare the magnitude of the magnetic field at points 1, 2 and 3 during this time interval. When computing magnetic fields, you may treat all wires as being infinitely long. This means that when calculating the magnetic field at points 1 and 3 you may use the corresponding formula that gives the magnetic field from an infinitely long wire. O(a) O(b) O(c) O(d) O(e) This study source was downloaded by 100000843892771 from CourseHero.com on 04-26-2024 08:21:43 GMT -05:010 hops /courses physics.illinois.edu'phys212/sp2018practice/practice.pllexam3/sp10 https.www.coursehero.com/file/30488221/Spring-2010-Physics-212-Hour-Exam-3pdf7 4/202018 Spring 2010 Physics 212 Hour Exam 3 a. 3 = 321=131 b. 181=2181=181 c. 181=21/2131=131 d 181=4131=181 e. 181=4/2131=181