University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

396

Pages:

5
Author:

Jabin F.

Normal Incidence (Di=0) (It is a special case of oblique incidence) Normal (Bi=0) [(\frac{E_{er}}{E_{oi}}){1}=(\frac{E{or}}{E_{ot}})=(\frac{n_{1}-n_{2}}{n_{i}+n_{2}y=0}] [|\frac{E_{0\gamma}}{E_{ei}}|_{1}=(\frac{E_{0\gamma}}{E_{0i}}){11}=\frac{n{1}-n_{2}}{n_{1}+n_{2}}] [(\frac{E_{et}}{E_{ei}}){1}=(\frac{E{et}}{E_{ei}}){11}=\frac{e^{4m{1}}}{(m_{1}+n_{2})}] An EM wave incident onto water from air normally, the change in phase of the reflected magnetic field is: (a) 0° (b) π (c) π (d) 3π/2 Solution: m1 = 1, m2 = 1.33 [(\frac{E_{or}}{E_{ot}}){ll}=(\frac{E{or}}{E_{ot}\dot{i}})_{L}=(\frac{1-8.33}{1+1.33})=-ve] Er suffers a phase change of π. Hr does not suffer a phase change. Page-80 UP 643 Option (a), (b), and (c) are right. Que. 9 Page 80 [I_{p}=\frac{I_{o}}{2}] (i) [I_{q}=-\frac{I_{o}}{2}cos^{2}wt] (ii) If angle between P and Q is θ, then angle between Q and R is (90-θ) So [I_{R}=I_{o}cos^{2}wt(cos^{2}(90-wt))] [=\frac{I_{o}}{2}sin^{2}(2wt)] wt=0 ⇒ IR=0 wt=π/4 ⇒ IR=I₀/8 wt=π/2 ⇒ IR=0 (Option b) is right. Page-82 [E_{0}(\hat{x}+e^{i\delta}\hat{y})exp{i(kz-wt)}] Phase diff. = δ ① If δ = 0, π, 2π, 3π... (m=0,1,2,3...) then → P.P. ② If δ = π/2, 3π/2, 5π/4... = (m+1)π/2 m=0,1,2... then → C.P ③ δ ≠ mπ and δ ≠ (2n+1)π/2 m=0, 1, 2, 3 then Light → E.P ④ P.P. → sin(nt) E.P. → 3E.F not vanished E.F vanished. For P.P. E.F vanishes (ω₀/2π) = ω₀/π in one second.