Electromagnetic Waves: Problem Solutions, Exams of Engineering Physics

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1.3 Part c

Write down the Poynting vector as well as the intensity ( the time averaged magnitude of S )

S =

μ 0 E^ ×^ B = 6. 64 W m^2

cos^2 (ky − ωt)ˆy

I = 〈|S|〉 = 6. 64

W

m^2

T

∫ T

0

cos^2 (ky − ωt) = 3. 32

W

m^2 I = 1 2

 0 cE 02 = 3. 318 W m^2

Both formulas for I agree.

1.4 Part d

If the wave strikes a perfectly reflecting mirror ( A = 10m^2 , m = 20kg ) square-on, what would be the acceleration of the mirror?

P = I

c F = 2 P A a = F m

=^2 IA

mc

= 1. 11 × 10 −^8 m s^2

yˆ

The force is multiplied by 2 as compared to a black surface because the light changes from having momentum in the +y direction to the −y direction rather than +y direction to none at all.

2 Problem 4

2.1 Part a

The first ray gets a π phase shift upon reflection by the oil. The second ray gets one upon reflection by the water. It also travels an extra 2t distance.

φ 1 = π φ 2 = π + 2t 2 π λoil ∆φ =

4 πtnoil λ = 2πm λ =

2 tnoil m

This is vacuum wavelength for constructive interference.

2.2 Part b

The ray that travels straight through picks up a phase from travelling a distance t through the oil while the second ray gets phase from the 3t of oil as well as the reflection from the oil water surface. It does not get any from the oil air reflection.

φ 1 = t^2 π λoil φ 2 = π +

3 t 2 π λoil ∆φ = 2 t^2 π λoil

• π = 2mπ 4 noilt λ

• 1 = 2 m

λ = (^24) mtn −oil 1 =^2 tnoil m − (^12)

We could also see this by replacing m in the formula from a. This is because a constructive interference for part a would be destructive for part b and vice versa.

2.3 Part c

First we find the angle of refraction

nair sin π 4

= noil sin θ

sin θ = √nair 2 noil

t

m

2 n^2 oil − n^2 air = 388 μm