The instantaneous Poynting vector.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Fresnel lens: You would like to design a 25 mm diameter blazed Fresnel zone plate with a first-order power of +1.5 diopters. What is the lithography requirement (resolution required) for making this lens that is designed for 550 nm? Express your answer in units of μm to one decimal point. Fresnel lens: What would the power of the first diffracted order of this lens be at wavelength of 400 nm? Express your answer in diopters to one decimal point. Eye: A person with myopic eyes has a far point of 15 cm. What power contact lenses does she need to correct her version to a standard far point at infinity? Give your answer in diopter to one decimal point.

Paraxial design of a field flattener. Imagine your optical system has Petzal curvature of the field with radius p. In Module 1 of Course 1, a homework problem asked you to derive the paraxial focus shift along the axis when a slab of glass was inserted in a converging cone of rays. Find or re-derive that result, then use it to calculate the paraxial radius of curvature of a field flattener of refractive index n that will correct the observed Petzval. Assume that the side of the flattener facing the image plane is plano. What is the required radius of the plano-convex field flattener? (p written as rho )

3.37(a) Five free electrons exist in a three-dimensional infinite potential well with all three widths equal to \( a = 12 \, \text{Å} \). Determine the Fermi energy level at \( T = 0 \, \text{K} \). (b) Repeat part (a) for 13 electrons. Book: Semiconductor Physics and Devices 4th ed, NeamanChapter-3Please expert answer only. don't give gpt-generated answers, & please clear the concept of quantum states for determining nx, ny, nz to determine E, as I don't have much idea about that topic.

Answer 2

Question

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Answer 6

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Answer 7

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

Answer 8

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Answer 13

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Answer 14

(b)

To determine

The time averaged Poynting vector.

Answer 15

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Answer 20

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Answer 21

(c)

To determine

The instantaneous and time average Poynting vector.

Answer 22

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.