Embed Size (px)
Citation preview
Plane Waves Part – II
1. For an electromagnetic wave incident from one medium to a second
medium, total reflection takes place when
(a) The angle of incidence is equal to the Brewster angle with E field
perpendicular to the plane of incidence.
(b) The angle of incidence is equal to the Brewster angle with E field
parallel to the plane of incidence.
(c) The angle of incidence is equal to the critical angle with the wave
moving from the denser medium to a rarer medium
(d) The angle of incidence is equal to the critical angle with the wave
moving from a rarer medium to a denser medium
[GATE 1987: 2 Marks]
Soln. For an electromagnetic wave incident from one medium to a second
medium, total (internal) reflection takes place when wave moves
from a denser to rarer medium and angle of incidence should be
greater than or equal to the critical angle.
The critical angle 𝜽𝒄,
𝐬𝐢𝐧(𝜽𝒄) = √𝝐𝟐
𝝐𝟏
𝐬𝐢𝐧(𝜽𝒊) ≥ √𝝐𝟐
𝝐𝟏
Where 𝜽𝒊 is the angle of incidence?
As 𝐬𝐢𝐧𝜽𝒊 𝒐𝒓 𝐬𝐢𝐧𝜽𝒄 can not be greater than 1, 𝝐𝟏 > 𝝐𝟐
Option (c)
2. In a good conductor the phase relation between the tangential components
of electric field Et and the magnetic field Ht is as follows
(a) Et and Ht are in phase
(b) Et and Ht are out of phase
(c) Ht leads Et by 900
(d) Et leads Ht by 450
[GATE 1988: 2 Marks]
Soln. For a good conductor 𝝈 ≫ 𝝎𝝐 Intrinsic impedance
𝜼 =𝑬
𝑯= √
𝒋𝝎𝝁
𝝈 + 𝒋𝝎𝝐
𝜼 = √𝝎𝝁
𝝈 ∠𝟒𝟓𝟎
�⃗⃗� 𝒍𝒆𝒂𝒅𝒔 �⃗⃗⃗� 𝒃𝒚 𝟒𝟓𝟎
Option (d)
3. The skin – depth of copper at a frequency of 3 GHz is 1 micron (10-6
meter). At 12 GHz, for a non – magnetic conductor whose conductivity is
1/9 times that of copper, the skin – depth would be
(a) √9×4 𝑚𝑖𝑐𝑟𝑜𝑛𝑠
(b) √9/4 𝑚𝑖𝑐𝑟𝑜𝑛𝑠
(c) √4/9 𝑚𝑖𝑐𝑟𝑜𝑛𝑠
(d) 1
√9×4 𝑚𝑖𝑐𝑟𝑜𝑛𝑠
[GATE 1989: 2 Marks]
Soln. For a good conductor, skin depth 𝜹 =𝟏
√𝝅𝒇𝝁𝝈
For copper 𝜹𝟏 =𝟏
√𝝅𝒇𝟏𝝁𝝈𝟏= 𝟏𝟎−𝟔𝒎
𝒇𝟏 = 𝟑 𝑮𝑯𝒛
For non - magnetic with 𝝁 = 𝝁𝟎
𝝈𝟐 =𝝈𝟏
𝟗 𝒂𝒕 𝒇𝟐 = 𝟏𝟐 𝑮𝑯𝒛
𝜹𝟐 =𝟏
√𝝅𝒇𝟐𝝁𝝈𝟐
𝜹 ∝𝟏
√𝒇𝝈 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒔𝒂𝒎𝒆 𝝁
𝜹𝟐
𝜹𝟏= √
𝒇𝟏𝝈𝟏
𝒇𝟐𝝈𝟐= √
𝟑 𝝈𝟏
𝟏𝟐𝝈𝟏𝟗
= √𝟗
𝟒=
𝟑
𝟐
Option (b)
4. The electric field component of a uniform plane electromagnetic wave
propagating in the Y – direction in a lossless medium will satisfy the
equation
(a) 𝜕2𝐸𝑦
𝜕𝑦2 = 𝜇 ∈𝜕2𝐸𝑦
𝜕𝑡2
(b) 𝜕2𝐸𝑦
𝜕𝑥2 = 𝜇 ∈𝜕2𝐸𝑦
𝜕𝑡2
(c) 𝜕2𝐸𝑋
𝜕𝑦2 = 𝜇 ∈𝜕2𝐸𝑋
𝜕𝑡2
(d) √𝐸𝑋
2+𝐸𝑍2
√𝐻𝑋2+𝐻𝑍
2= √
𝜇
∈
[GATE 1991: 2 Marks]
Soln. In a uniform plane EM wave propagating in the y – direction, the
components of �⃗⃗� and �⃗⃗⃗� in the direction of propagation Y 𝑬𝒚, 𝑯𝒚 are
zero. �⃗⃗� and �⃗⃗⃗� should be function of Y and t satisfying second order
partial differential equation
𝜹𝟐𝑬𝒙
𝜹 𝒚𝟐= 𝝁𝝐
𝝏𝟐𝑬𝒙
𝝏𝒕𝟐
�⃗⃗� and �⃗⃗⃗� are related as
𝑬𝒛
𝑯𝒙= 𝜼 ,
𝑬𝒙
𝑯𝒛= −𝜼
𝑬
𝑯= √
𝑬𝒙𝟐 + 𝑬𝒛
𝟐
𝑯𝒙𝟐 + 𝑯𝒛
𝟐= 𝜼
𝜼 = 𝐈𝐧𝐭𝐫𝐢𝐧𝐬𝐢𝐜 𝐢𝐦𝐩𝐞𝐝𝐚𝐧𝐜𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐦𝐞𝐝𝐢𝐮𝐦
= 𝜼𝟎
= √𝝁
∈
Option (c) and (d)
5. A material is described by the following electrical parameters at a
frequency of 10 GHz 𝜎 = 106 𝑚ℎ𝑜
𝑚, 𝜇 = 𝜇0 𝑎𝑛𝑑 ∈ ∈0⁄ = 10. The
material at this frequency is considered to be (∈0=1
36𝜋×10−9 𝐹 𝑚⁄ )
(a) Good conductor
(b) Good dielectric
(c) Neither a good conductor nor a good dielectric
(d) Good magnetic material
[GATE 1993: 2 Marks]
Soln. For a good conductor 𝝈
𝝎∈≫ 𝟏
𝒇 = 𝟏𝟎 𝑮𝑯𝒛, 𝝈 = 𝟏𝟎𝟔 𝒎𝒉𝒐/𝒎
∈
∈𝟎= 𝟏𝟎 𝒐𝒓 ∈= 𝟏𝟎
𝟏
𝟑𝟔𝝅×𝟏𝟎𝟗
=𝟏𝟎−𝟖
𝟑𝟔𝝅
𝝈
𝝎∈=
𝟏𝟎𝟔 ×𝟑𝟔𝝅
𝟐𝝅×𝟏𝟎×𝟏𝟎𝟗×𝟏𝟎−𝟖
=𝟏𝟎𝟔×𝟑𝟔𝝅×𝟏𝟎𝟖
𝟐𝝅×𝟏𝟎𝟏𝟎
= 𝟏𝟖×𝟏𝟎𝟒 ≫ 𝟏
Option (a)
6. A plane wave is incident normally on a perfect conductor as shown in
figure. Here 𝐸𝑥𝑖 , 𝐻𝑦
𝑖 and 𝑃𝑖⃗⃗ ⃗ are electric field, magnetic field and
Poynting vector respectively for the incident wave. The reflected wave
should have
𝐸𝑋𝑖
𝐻𝑌𝑖
𝑃𝑖 Z
X
(a) 𝐸𝑥𝑟 = −𝐸𝑥
𝑖
(b) 𝐻𝑦𝑟 = −𝐻𝑦
𝑖
(c) 𝑃𝑟⃗⃗ ⃗⃗ = −𝑃𝑖⃗⃗ ⃗
(d) 𝐸𝑥𝑟 = −𝐸𝑋
𝑖
[GATE 1993: 2 Marks]
Soln. The tangential component of E is continuous at the surface. That is it
is the same just outside the surface as it is just inside the surface.
As E is zero within a perfect conductor, tangential component just
outside the conductor at 𝒛 = 𝟎,= 𝑬𝒙𝒊 + 𝑬𝒙
𝒓
Tangential component just inside the conductor at 𝒛 = 𝟎+= 𝟎
𝑬𝒙𝒓 = −𝑬𝒙
𝒊
The amplitude of 𝑬𝒙𝒊 is reversed on reflection, but 𝑯𝒙
𝒓 = 𝑯𝒙𝒊
�⃗⃗� 𝒓 = −�⃗⃗� 𝒊
The average power flow is zero indicating a standing wave in the
incident – reflected medium
Option (a) & (c)
7. A uniform plane wave in air is normally incident on an infinitely thick
slab. If the refractive index of glass slab is 1.5, then the percentage of the
incident power that is reflected from the air – glass interface is
(a) 0%
(b) 4%
(c) 20%
(d) 100%
[GATE 1996: 2 Marks]
Soln. Medium 1 Medium 2
𝒏𝟏 𝒏𝟐 = 𝟏. 𝟓
𝝁𝟏 = 𝝁𝟎 𝝁𝟐 = 𝝁𝟎
∈𝟏=∈𝟎 ∈𝟐=∈𝟎∈𝒓
If 𝒏𝟏 , 𝒏𝟐 are refractive indices and 𝒗𝟏 , 𝒗𝟐 are the velocities
𝒏𝟏
𝒏𝟐=
𝒗𝟐
𝒗𝟏=
√𝝁𝟏𝝐𝟏
√𝝁𝟐𝝐𝟐
=𝟏
𝟏. 𝟓
= √∈𝟏
∈𝟐 𝒇𝒐𝒓 𝝁𝟏 = 𝝁𝟐 = 𝝁𝟎
√𝝐𝟏
𝝐𝟐=
𝟏
𝟏. 𝟓=
𝟐
𝟑
Reflection coefficient, 𝑬𝒓
𝑬𝒊=
𝜼𝟐−𝜼𝟏
𝜼𝟐+𝜼𝟏 𝜼𝟏 𝒂𝒏𝒅 𝜼𝟐 are the intrinsic
impedances of medium 1 and medium 2 respectively
𝜼𝟏 = √𝝁𝟏
𝝐𝟏 , 𝜼𝟐 = √
𝝁𝟐
𝝐𝟐
𝝁𝟏 = 𝝁𝟐 = 𝝁𝟎
𝑬𝒓
𝑬𝒊=
√∈𝟏− √∈𝟐
√∈𝟏+ √∈𝟐
𝑬𝒓
𝑬𝒊=
√∈𝟏∈𝟐
−𝟏
√∈𝟏∈𝟐
+𝟏=
𝟐
𝟑−𝟏
𝟐
𝟑+𝟏
=−
𝟏
𝟑𝟓
𝟑
= −𝟏
𝟓
𝑷𝒓
𝑷𝒊=
|𝑬𝒓|𝟐
|𝑬𝒊|𝟐 =
𝟏
𝟐𝟓= 𝟒%
Option (b)
8. Some unknown material has a conductivity of 106 mho/m and a
permeability of 4𝜋×10−7 𝐻/𝑚. The skin depth for the material at 1 GHz
is
(a) 15.9 𝜇𝑚
(b) 20.9 𝜇𝑚
(c) 25.9 𝜇𝑚
(d) 30.9 𝜇𝑚
[GATE 1996: 2 Marks]
Soln. 𝝈 = 𝟏𝟎𝟔𝒎𝒉𝒐/𝒎
𝝁 = 𝝁𝟎 = 𝟒𝝅×𝟏𝟎−𝟕 𝑯/𝒎
𝒇 = 𝟏 𝑮𝑯𝒛
𝒔𝒌𝒊𝒏 𝒅𝒆𝒑𝒕𝒉 𝜹 =𝟏
√𝝅𝒇𝝁𝝈=
𝟏
√𝝅×𝟏𝟎𝟗×𝟒𝝅×𝟏𝟎−𝟕×𝟏𝟎𝟔
=𝟏
𝟐𝝅×𝟏𝟎𝟒 𝒎 =𝟓𝟎
𝝅 𝝁𝒎
= 𝟏𝟓. 𝟗 𝝁𝒎
Option (a)
9. A uniform plane wave in air impinges at 450 angle on a lossless dielectric
material with dielectric constant 휀𝑟. The transmitted wave propagates in a
300 direction with respect to the normal. The value of 휀𝑟 is
(a) 1.5
(b) √1.5
(c) 2
(d) √2
[GATE 2000: 2 Marks]
Soln. At the interface between air and lossless dielectric angle of incidence
𝜽𝒊 = 𝟒𝟓𝟎 angle of refraction 𝜽𝒓 = 𝟑𝟎𝟎
450
300
According to Snell’s law
𝐬𝐢𝐧𝜽𝟐
𝐬𝐢𝐧𝜽𝟏=
𝒗𝟐
𝒗𝟏
𝒗𝟐 is the velocity of wave in medium 2
𝒗𝟏 is the velocity of wave in medium 1
𝒗𝟐 = 𝟏
√𝝁𝟐 ∈𝟐
, 𝒗𝟏 = 𝟏
√𝝁𝟏 ∈𝟏
𝝁𝟏 = 𝝁𝟐 = 𝝁𝟎 , ∈𝟐=∈𝒓∈𝟏
𝐬𝐢𝐧𝜽𝟐
𝐬𝐢𝐧𝜽𝟏=
𝐬𝐢𝐧 𝟑𝟎𝟎
𝐬𝐢𝐧 𝟒𝟓𝟎=
√𝝁𝟏 ∈𝟏
√𝝁𝟐 ∈𝟐
𝐬𝐢𝐧𝟑𝟎𝟎
𝐬𝐢𝐧𝟒𝟓𝟎=
𝟏×√𝟐
𝟐×𝟏=
𝟏
√∈𝒓
𝟏
√𝟐=
𝟏
√∈𝒓
Option (d)
10. A uniform plane electromagnetic wave incident normally on a plane
surface of a dielectric material is reflected with a VSWS of 3. What is the
percentage of incident power that is reflected?
(a) 10%
(b) 25%
(c) 50%
(d) 75%
[GATE 2001: 2 Marks]
Soln. VSWR = 3
Let S = 3
𝑺 =𝟏+|𝑲|
𝟏−|𝑲| , 𝒌 𝒊𝒔 𝒕𝒉𝒆 𝒓𝒆𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒏𝒆𝒕
|𝑲| =𝑺−𝟏
𝑺+𝟏=
𝟑−𝟏
𝟑+𝟏=
𝟏
𝟐
𝑬𝒓
𝑬𝒊= 𝑲
𝑷𝒓
𝑷𝒊=
|𝑬𝒓|𝟐
|𝑬𝒊|𝟐 = 𝑲𝟐 =
𝟏
𝟒
= 25%
Option (b)
11. A plane wave is characterized by �⃗� = (0.5�̂� + �̂�𝑒𝑗𝜋/2)𝑒𝑗𝜔𝑡−𝑗𝑘𝑧. This
wave is
(a) Linearly polarized
(b) Circularly polarized
(c) Elliptically polarized
(d) Un polarized
[GATE 2002: 2 Marks]
Soln. The wave �⃗⃗� = (𝟎. 𝟓�̂� + �̂� 𝒆𝒋𝝅/𝟐)𝒆𝒋𝝎𝒕−𝒋𝒌𝒛 represents a wave travelling
in the positive Z direction. Taking the real part of �⃗⃗�
𝑬𝒙(𝒛, 𝒕) = 𝟎. 𝟓 𝐜𝐨𝐬(𝝎𝒕 − 𝒌𝒛)
𝑬𝒚(𝒛, 𝒕) = 𝐜𝐨𝐬 (𝝎𝒕 +𝝅
𝟐− 𝒌𝒛)
= −𝐬𝐢𝐧(𝝎𝒕 − 𝒌𝒛)
(𝑬𝒙
𝟎. 𝟓)𝟐
+ (𝑬𝒚
𝟏)
𝟐
= 𝟏
This is the equation of an ellipse. The wave is elliptically polarized
Option (c)
12. Distilled water at 250 C is characterized by 𝜎 = 1.7×10−4 𝑚ℎ𝑜
𝑚 𝑎𝑛𝑑 ∈=
78 ∈0 at a frequency of 3 GHz. Its loss tangent tan 𝛿 is
(a) 1.3×10−5
(b) 1.3×10−3
(c) 1.7×10−4 78⁄
(d) 1.7×10−4(78 ∈0)
[GATE 2002: 2 Marks]
Soln. For distilled water 𝝈 = 𝟏. 𝟕×𝟏𝟎−𝟒 𝒎𝒉𝒐/𝒎
∈= 𝟕𝟖 ∈𝟎 , 𝒇 = 𝟑 𝑮𝑯𝒛 , ∈𝟎=𝟏𝟎−𝟗
𝟑𝟔𝝅
Loss tangent = 𝐭𝐚𝐧𝜹 =𝝈
𝝎∈
𝝈
𝝎 ∈=
𝟏. 𝟕×𝟏𝟎−𝟒×𝟑𝟔𝝅×𝟏𝟎𝟗
𝟐𝝅×𝟑×𝟏𝟎𝟗×𝟕𝟖
= 𝟎. 𝟏𝟑𝟎×𝟏𝟎−𝟒
≅ 𝟏. 𝟑×𝟏𝟎−𝟓
𝐽 𝑑
𝐽
𝜹
𝑱 = conduction current density = 𝝈�⃗⃗�
𝑱 𝒅 = displacement current density = 𝒋𝝎 ∈ �⃗⃗�
Option (a)
13. Medium 1 has the electrical permittivity 휀1 = 1.5 휀0 𝑓𝑎𝑟𝑎𝑑/𝑚 and
occupies the region to left of 𝑥 = 0 plane. Medium 2 has the electrical
permittivity 휀2 = 2.5 휀0 farad / m and occupies the region to the right of
𝑥 = 0 plane. If E1 in medium 1 is 𝐸1 = (2 𝑢𝑥 − 3 𝑢𝑦 + 1 𝑢𝑧) volt/m, the
E2 in medium 2 is
(a) (2 𝑢𝑋 − 7.5 𝑢𝑌 + 2.5 𝑢𝑍) 𝑣𝑜𝑙𝑡 𝑚⁄
(b) (2 𝑢𝑋 − 2 𝑢𝑌 + 0.6 𝑢𝑍) 𝑣𝑜𝑙𝑡 𝑚⁄
(c) (1.2 𝑢𝑋 − 3 𝑢𝑌 + 1 𝑢𝑍) 𝑣𝑜𝑙𝑡 𝑚⁄
(d) (1.2 𝑢𝑋 − 2 𝑢𝑌 + 0.6 𝑢𝑍) 𝑣𝑜𝑙𝑡 𝑚⁄
[GATE 2003: 2 Marks]
Soln. The interface between medium 1 and 2 is x = 0 or y – z plane.
𝑋 < 0
∈1= 1.5 ∈0
𝑋 = 0
𝑋 > 0
∈2= 2.5 ∈0
𝑌 − 𝑍 𝑝𝑙𝑎𝑛𝑒
1 2
𝑬𝟏 = (𝟐 𝒖𝒙 − 𝟑 𝒖𝒚 + 𝟏 𝒖𝒛) 𝒗𝒐𝒍𝒕/𝒎
The y and z components of E1 are same as the y and z components of
E2 as the tangential components of E are continuous.
𝑬𝒕𝟏 = 𝑬𝒕𝟐
The normal components of D are continuous
𝑫𝑵𝟏= 𝑫𝑵𝟐
∈𝟏 𝑬𝑵𝟏=∈𝟐 𝑬𝑵𝟐
∈𝟏 𝑬𝑿𝟏=∈𝟐 𝑬𝑿𝟐
𝒐𝒓 𝑬𝑿𝟐=
∈𝟏 𝑬𝑿𝟏
∈𝟐=
𝟏. 𝟓 ∈𝟎×𝟐
𝟐. 𝟓 ∈𝟎= 𝟏. 𝟐
�⃗⃗� 𝟏 = (𝟏. 𝟐 𝒖𝒙 − 𝟑 𝒖𝒚 + 𝟏 𝒖𝒛) 𝒗𝒐𝒍𝒕/𝒎
Option (c)
14. A uniform plane wave traveling in air is incident on the plane boundary
between air and another dielectric medium with 휀𝑟 = 4. The reflection
coefficient for the normal incidence is
(a) Zero
(b) 0.5 ∠1800
(c) 0.333 ∠00
(d) 0.333 ∠1800
[GATE 2003: 2 Marks]
Soln. For normal incidence of a uniform plane wave at air dielectric
interface, reflection coefficient 𝑲, ∈𝒓= 𝟒
𝑲 =𝑬𝒓
𝑬𝒊=
𝜼𝟐 − 𝜼𝟏
𝜼𝟐 + 𝜼𝟏
=
𝜼𝟐
𝜼𝟏− 𝟏
𝜼𝟐
𝜼𝟏+ 𝟏
𝝁𝟏 = 𝝁𝟐 = 𝝁𝟎 , 𝜼𝟐 = √𝝁𝟎
∈𝟐
𝜼𝟏 = √𝝁𝟎
∈𝟏
𝑲 =√
∈𝟏
∈𝟐− 𝟏
√∈𝟏
∈𝟐+ 𝟏
Where, ∈𝟏=∈𝟎 and ∈𝟐=∈𝒓∈𝟎
𝑲 =√
𝟏∈𝒓
− 𝟏
√𝟏∈𝒓
+ 𝟏
=
𝟏𝟐
− 𝟏
𝟏𝟐
+ 𝟏= −
𝟏
𝟑
𝑲 = −𝟏
𝟑= 𝟎. 𝟑𝟑𝟑∠𝟏𝟖𝟎𝟎
Option (d)
15. If the electric field intensity associated with a uniform plane
electromagnetic wave traveling in a perfect dielectric medium is given by
𝐸(𝑧, 𝑡) = 10 cos(2𝜋×107𝑡 − 0.1 𝜋𝑧) 𝑣𝑜𝑙𝑡/𝑚, the velocity of the
traveling wave is
(a) 3.00×108 𝑚/𝑠𝑒𝑐
(b) 2.00×108 𝑚/𝑠𝑒𝑐
(c) 6.28×107 𝑚/𝑠𝑒𝑐
(d) 2.00×107 𝑚/𝑠𝑒𝑐
[GATE 2003: 2 Marks]
Soln. For a uniform plane wave in a perfect dielectric medium 𝝈 = 𝟎
𝑬(𝒛, 𝒕) = 𝟏𝟎 𝐜𝐨𝐬(𝟐𝝅×𝟏𝟎𝟕𝒕 − 𝟎. 𝟏𝝅𝒛) 𝒗/𝒎
The E field represents wave travelling in the positive z direction with
𝝎 = 𝟐𝝅×𝟏𝟎𝟕 𝒓𝒂𝒅/ 𝐬𝐞𝐜 𝜷 = 𝟎. 𝟏𝝅 𝒓𝒂𝒅/𝒔𝒆𝒄
Velocity of propagation 𝑽 =𝝎
𝜷
=𝟐𝝅×𝟏𝟎𝟕
𝟎.𝟏𝝅
= 𝟐×𝟏𝟎𝟖 𝒎/𝒔𝒆𝒄
Option (b)
16. A plane electromagnetic wave propagating in free space is incident
normally on a large slab of loss – less, non – magnetic, dielectric material
with 휀 > 휀0. Maxima and minima are observed when the electric field is
measured in front of the slab. The maximum electric field is found to be 5
times the minimum field. The intrinsic impedance of the medium should
be
(a) 120 𝜋 Ω
(b) 60 𝜋 Ω
(c) 600 𝜋 Ω
(d) 24 𝜋 Ω
[GATE 2004: Marks]
Soln. 𝑬𝒎𝒂𝒙 = 𝟓 𝑬𝒎𝒊𝒏 𝒊𝒏 𝒎𝒆𝒅𝒊𝒖𝒎 𝟏
𝑽𝑺𝑾𝑹,𝑺 =𝑬𝒎𝒂𝒙
𝑬𝒎𝒊𝒏= 𝟓
Reflection coefficient |𝑲| =𝑺−𝟏
𝑺+𝟏=
𝟓−𝟏
𝟓+𝟏=
𝟐
𝟑
1 2
free space loss less
𝜎 = 0, 𝜇 = 𝜇0 𝜎 = 0
∈ =∈0 Non magnetic
Incident wave
Reflected wave
𝜇 = 𝜇0
dielectric
∈ > ∈0
𝑲 =
𝜼𝟐𝜼𝟏
− 𝟏
𝜼𝟐𝜼𝟏
+ 𝟏=
𝟐
𝟑
𝟑𝜼𝟐
𝜼𝟏− 𝟑 =
𝟐𝜼𝟐
𝜼𝟏+ 𝟐
𝜼𝟐
𝜼𝟏= 𝟓 , 𝜼𝟐 = 𝟓𝜼𝟏
𝜼𝟏 = √𝝁𝟎
∈𝟎= √𝟒𝝅×𝟏𝟎−𝟕×𝟑𝟔𝝅×𝟏𝟎𝟗
= (𝟏𝟐𝟎𝝅)𝛀
𝜼𝟐 = 𝟓×𝟏𝟐𝟎𝝅
= 𝟔𝟎𝟎 𝝅 𝛀
Option (c)
17. A medium is divided into regions I and II about x = 0 plane, as shown in
the figure below. An electromagnetic wave with electric field
�⃗� 1 = 4�̂�𝑥 + 3�̂�𝑦 + 5�̂�𝑧 is incident normally on the interface from region
– I. The electric filed E2 in region – II at the interface is
𝐸1 𝐸2
Region I Region II
𝜎1 = 0 , 𝜇1 = 𝜇0 𝜎2 = 0 , 𝜇2 = 𝜇0
∈𝑟1= 3 ∈𝑟2
= 4
𝑋 = 0
(a) 𝐸2 = 𝐸1
(b) 4 �̂�𝑥 + 0.7 �̂�𝑦 − 1.25 �̂�𝑧
(c) 3 �̂�𝑥 + 3 �̂�𝑦 + 5 �̂�𝑧
(d) −3 �̂�𝑥 + 3 �̂�𝑦 + 5 �̂�𝑧
[GATE 2006: 2 Marks]
Soln. 𝑬𝟏 = 𝟒�̂�𝒙 + 𝟑�̂�𝒚 + 𝟓�̂�𝒛
The Y and Z components of E1 are same as the Y and Z components
of E2 as the tangential components of E are continuous at the
boundary.
𝑬𝒕𝟏 = 𝑬𝒕𝟐
𝑬𝒚𝟐= 𝑬𝒚𝟏
, 𝑬𝒛𝟐= 𝑬𝒛𝟏
The normal components of D is continuous at the boundary 𝑫𝒏𝟏=
𝑫𝒏𝟐
𝝐𝟏𝑬𝒙𝟏= 𝝐𝟐𝑬𝒙𝟐
𝒐𝒓 𝑬𝒙𝟐=
𝝐𝟏𝑬𝒙𝟏
∈𝟐=
𝟑
𝟒×
𝟒
𝟏
𝑬𝟐 = 𝟑�̂�𝒙 + 𝟑�̂�𝒚 + 𝟓�̂�𝒛
Option (c)
18. When a plane wave traveling in free – space is incident normally on a
medium having 휀𝑟 = 4.0, the fraction of power transmitted into the
medium is given by
(a) 8/9
(b) 1/2
(c) 1/3
(d) 5/6
[GATE 2006: 2 Marks]
Soln. Normal incidence from free space on a medium is shown in figure
𝑬𝒕
𝑬𝒊=
𝟐𝜼𝟐
𝜼𝟐 + 𝜼𝟏=
𝟐
𝟏 +𝜼𝟏
𝜼𝟐
1 2
free space
𝜇 = 𝜇0
∈𝑟= 4
𝜇2 = 𝜇0
∈ = ∈0
𝜂1 = 𝜂0 𝐸𝑖
𝑃𝑖 𝑃𝑡
𝐸𝑡
𝐸𝑟
𝑃𝑟
𝜼𝟏
𝜼𝟐= √
𝝁𝟐 ∈𝟐
𝝁𝟏 ∈𝟏 𝝁𝟏 = 𝝁𝟐 = 𝝁𝟎
= √∈𝒓= 𝟐
𝑬𝒕
𝑬𝒊=
𝟐
𝟏 + 𝟐=
𝟐
𝟑
𝑷𝒊 = 𝑰𝒏𝒄𝒊𝒅𝒆𝒏𝒕 𝒑𝒐𝒘𝒆𝒓 =𝑬𝒊
𝟐
𝜼𝟏
𝑷𝒕 = 𝑻𝒓𝒂𝒏𝒔𝒎𝒊𝒕𝒕𝒆𝒅 𝒑𝒐𝒘𝒆𝒓 =𝑬𝒕
𝟐
𝜼𝟐
𝑷𝒕
𝑷𝒊=
𝑬𝒕𝟐
𝑬𝒊𝟐 𝜼𝟏
𝜼𝟐=
𝟒
𝟗×𝟐 =
𝟖
𝟗
Option (a)
19. The �⃗⃗� field (in A/m) of a plane wave propagating in free space is given
by �⃗⃗� = 𝑥5√3
𝜂0cos(𝜔𝑡 − 𝛽𝑧) + �̂�
5
𝜂0sin (𝜔𝑡 − 𝛽𝑧 +
𝜋
2). The time average
power flow density in Watts is
(a) 𝜂0
100
(b) 100
𝜂0
(c) 50𝜂02
(d) 50
𝜂0
[GATE 2007: Marks]
Soln.
�⃗⃗⃗� = �̂�𝟓√𝟑
𝜼𝟎𝐜𝐨𝐬(𝝎𝒕 − 𝜷𝒛) + �̂�
𝟓
𝜼𝟎𝐬𝐢𝐧 (𝝎𝒕 − 𝜷𝒛 +
𝝅
𝟐)
𝑯𝒙 =𝟓√𝟑
𝜼𝟎 , 𝑯𝒚 =
𝟓
𝜼𝟎
𝑯𝑻 = √𝑯𝒙𝟐 + 𝑯𝒚
𝟐 = √(𝟓√𝟑)𝟐
𝜼𝟎𝟐 +
𝟓𝟐
𝜼𝟎𝟐
=𝟓
𝜼𝟎
√(√𝟑)𝟐+ 𝟏
=𝟓
𝜼𝟎√𝟒
=𝟏𝟎
𝜼𝟎
Time average power flow density
𝑷𝒂𝒗𝒈 =𝟏
𝟐𝜼𝟎𝑯𝑻
𝟐
=𝟏
𝟐𝜼𝟎
𝟏𝟎𝟎
𝜼𝟎𝟐
=𝟓𝟎
𝜼𝟎 𝒘𝒂𝒕𝒕𝒔
Option (d)
20. A plane wave having the electric field component
�⃗� 𝑖 = 24 cos(3×108𝑡 − 𝛽𝑦) �̂�𝑧 𝑉/𝑚 and traveling in free space is
incident normally on a lossless medium with 𝜇 = 𝜇0 and 휀 = 9 휀0 which
occupies the region 𝑦 ≥ 0. The reflected magnetic field component is
given by
(a) 1
10𝜋cos(3×108𝑡 + 𝑦) �̂�𝑥 𝐴/𝑚
(b) 1
20𝜋cos(3×108𝑡 + 𝑦) �̂�𝑥 𝐴/𝑚
(c) −1
20𝜋cos(3×108𝑡 + 𝑦) �̂�𝑥 𝐴/𝑚
(d) −1
10𝜋cos(3×108𝑡 + 𝑦) �̂�𝑥 𝐴/𝑚
[GATE 2010: 2 Marks]
Soln. 𝑬𝒊𝒛 = 𝟐𝟒𝐜𝐨𝐬(𝟑×𝟏𝟎𝟖𝒕 − 𝜷𝒚) 𝒗/𝒎
𝝎 = 𝟑×𝟏𝟎𝟖 𝒓𝒂𝒅/𝒔𝒆𝒄
𝜷 =𝝎
𝑽
𝑽 = 𝑽𝟎 𝒇𝒐𝒓 𝒇𝒓𝒆𝒆 𝒔𝒑𝒂𝒄𝒆 = 𝟑×𝟏𝟎𝟖 𝒎/𝒔𝒆𝒄
𝜷 = 𝟏 𝒓/𝒎
𝜼𝟏 = 𝜼𝟎 = 𝟏𝟐𝟎𝝅 =𝑬𝒊𝒛
𝑯𝒊𝒙
𝑯𝒊𝒙 =𝟐𝟒𝐜𝐨𝐬(𝟑×𝟏𝟎𝟖𝒕 − 𝒚)
𝟏𝟐𝟎𝝅=
𝐜𝐨𝐬(𝟑×𝟏𝟎𝟖𝒕 − 𝒚)
𝟓𝝅
𝑯𝒓
𝑯𝒊=
𝜼𝟏 − 𝜼𝟐
𝜼𝟏 + 𝜼𝟐=
𝜼𝟏
𝜼𝟐− 𝟏
𝜼𝟏
𝜼𝟐+ 𝟏
𝜼𝟏
𝜼𝟐=
√𝝁𝟐 ∈𝟐
√𝝁𝟏 ∈𝟏
𝝁𝟐 = 𝝁𝟏 = 𝝁𝟎
𝜼𝟏
𝜼𝟐=
√∈𝟐
√∈𝟏
= √𝟗 ∈𝟎
∈𝟎= 𝟑
𝑯𝒓
𝑯𝒊=
𝟑 − 𝟏
𝟑 + 𝟏=
𝟏
𝟐
�⃗⃗⃗� 𝒓 =𝟏
𝟐𝐜𝐨𝐬(𝟑×𝟏𝟎𝟖𝒕 + 𝟏𝒚) �⃗⃗� 𝒙 𝑨/𝒎
�⃗⃗⃗� 𝒓 is reflected wave which travels in negative Y direction
Option (a)
21. The electric and magnetic fields for a TEM wave of frequency 14 GHz in
a homogeneous medium of relative permittivity 휀𝑟 and relative
permeability 𝜇𝑟 = 1 are given by
�⃗� = 𝐸𝑃 𝑒𝑗(𝜔𝑡−280𝜋𝑦) �̂�𝑧 𝑉 / 𝑚
�⃗⃗� = 3 𝑒𝑗(𝜔𝑡−280𝜋𝑦) �̂�𝑥 𝑉 / 𝑚
Assuming the speed of light in free space to be 3×108 𝑚/𝑠, the intrinsic
impedance of free space to be 120 π, the relative permittivity 휀𝑟 of the
medium and the electric field amplitude 𝐸𝑃 are
(a) 휀𝑟 = 3, 𝐸𝑃 = 120𝜋
(b) 휀𝑟 = 3, 𝐸𝑃 = 360𝜋
(c) 휀𝑟 = 9, 𝐸𝑃 = 360𝜋
(d) 휀𝑟 = 9, 𝐸𝑃 = 120𝜋
[GATE 2011: 2 Marks]
Soln. �⃗⃗� 𝒂𝒏𝒅 �⃗⃗⃗� , 𝐭𝐡𝐞 𝐰𝐚𝐯𝐞 𝐢𝐬 𝐭𝐫𝐚𝐯𝐞𝐥𝐢𝐧𝐠 𝐢𝐧 𝐭𝐡𝐞 𝐘 − 𝐝𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧
Phase shift constant 𝜷 = 𝟐𝟖𝟎𝝅
𝜼 =𝑬
𝑯=
𝑬𝑷
𝟑
Velocity of wave 𝑽 =𝝎
𝜷
𝑽 =𝟐𝝅×𝟏𝟒×𝟏𝟎𝟗
𝟐𝟖𝟎𝝅
= 𝟏𝟎𝟖 𝒎/𝒔
𝑽 =𝟏
√𝝁 ∈=
𝟏
√𝝁𝟎 ∈𝟎
×𝟏
√𝝁𝒓 ∈𝒓
Velocity for free space 𝟏
√𝝁𝟎∈𝟎= 𝟑×𝟏𝟎𝟖 𝒎/𝒔𝒆𝒄
𝑽 = 𝟑×𝟏𝟎𝟖×𝟏
√𝝁𝒓∈𝒓
𝝁𝒓 = 𝟏
𝑽 =𝟑×𝟏𝟎𝟖
√∈𝒓= 𝟏𝟎𝟖 , ∈𝒓= 𝟗
𝜼 = √𝝁
∈= √
𝝁𝒓𝝁𝟎
∈𝒓∈𝟎= √
𝝁𝟎
∈𝟎×
𝟏
√∈𝒓
𝝁𝒓 = 𝟏
𝜼 =𝟏𝟐𝟎𝝅
√∈𝒓
𝑬𝑷
𝟑=
𝟏𝟐𝟎𝝅
√∈𝒓
𝑬𝑷 = 𝟏𝟐𝟎𝝅 𝑽/𝒎
Option (c)
22. If the electric field of a plane wave is
�⃗� (𝑧, 𝑡) = 𝑥3 cos(𝜔𝑡 − 𝑘𝑧 + 300) − �̂�4 sin(𝜔𝑡 − 𝑘𝑧 + 450) (𝑚𝑉/𝑚)
The polarization state of the plane wave is
(a) Left elliptical
(b) Left circular
(c) Right elliptical
(d) Right circular
[GATE 2014: 2 Mark]
Soln. Equation for electric field of plane wave is
�⃗⃗� (𝒛, 𝒕) = �̂�. 𝟑 𝐜𝐨𝐬(𝝎𝒕 − 𝒌𝒛 + 𝟑𝟎𝟎) − �̂�. 𝟒 𝐬𝐢𝐧(𝝎𝒕 − 𝒌𝒛 + 𝟒𝟓𝟎) (𝒎𝑽/
𝒎)
Here wave is propagation in +z direction. Above equation can be
written for z = 0
�⃗⃗� (𝟎, 𝒕) = �̂�. 𝟑 𝐜𝐨𝐬(𝝎𝒕 + 𝟑𝟎𝟎) − 𝒚. 𝟒 𝐬𝐢𝐧(𝝎𝒕 + 𝟒𝟓𝟎)
= �̂�. 𝟑 𝐜𝐨𝐬(𝝎𝒕 + 𝟑𝟎𝟎) + 𝒚. 𝟒 𝐬𝐢𝐧(𝝎𝒕 + 𝟒𝟓𝟎 + 𝟏𝟖𝟎𝟎)
𝑬(𝟎, 𝒕) = �̂�. 𝟑 𝐜𝐨𝐬(𝝎𝒕 + 𝟑𝟎𝟎) + �̂�. 𝟒 𝐜𝐨𝐬(𝝎𝒕 + 𝟏𝟑𝟓𝟎)
Ey component is leading Ex by 1050
As in the given figure the resultant E vector rotates in clockwise
direction as shown. Since the direction of propagation is +z direction.
If we look the wave in +z direction then it will look moving counter
clockwise or left hand elliptically polarized (|𝑬𝒙| ≠ |𝑬𝒚|)
𝑥
𝑦
𝑧
𝐸
Hint: For right handed coordinate system. In the given figure
𝑥
𝑦
𝑧 𝐸1
𝐸2
𝐸
E1 - amplitude of wave linearly polarized in x direction
E2 - amplitude of wave linearly polarized in y direction
E - Resultant vector
If E2 leads E1, then E vector moves (clockwise) as indicated
If E1 leads E2 then it moves in counter clockwise. To observe the wave
polarization, we have to look towards the direction of propagation of
wave.
(Refer: Antennas by Kraus)
23. The electric field of a plane wave propagation in a lossless non –
magnetic medium is given by the following expression
�⃗� (𝑧, 𝑡) = �̂�𝑥5 cos(2𝜋×109𝑡 + 𝛽𝑧) + �̂�𝑦3 cos (2𝜋×109𝑡 + 𝛽𝑧 −𝜋
2)
The type of the polarization is
(a) Right hand circular
(b) Left hand elliptical
(c) Right hand elliptical
(d) Linear
[GATE 2015: 2 Mark]
Soln. Given
�⃗⃗� (𝒛, 𝒕) = �̂�𝒙𝟓𝐜𝐨𝐬(𝟐𝝅×𝟏𝟎𝟗𝒕 + 𝜷𝒛) + �̂�𝒚𝟑𝐜𝐨𝐬 (𝟐𝝅×𝟏𝟎𝟗𝒕 + 𝜷𝒛 −𝝅
𝟐)
Let 𝝎 = 𝟐𝝅×𝟏𝟎𝟗
�⃗⃗� (𝒛, 𝒕) = �̂�𝒙𝟓𝐜𝐨𝐬(𝝎𝒕 + 𝜷𝒛) + �̂�𝒚𝟑𝐜𝐨𝐬 (𝝎𝒕 + 𝜷𝒛 −𝝅
𝟐)
Here, Ey is lagging Ex by 900
The resultant vector moves as shown (opposite to leading case) as
shown in the given figure
𝑧
𝐸𝑥
𝐸𝑦
𝐸
−𝑧
Direction of propagation as per the equation is –z direction
Thus, the resultant vector E moves in counter clock clockwise if we
look into –z direction
Thus, the wave is left hand elliptically polarized (since unequal
amplitude)
Option (b)
