EC2402 – OPTICAL COMMUNICATION AND NETWORKS
OPTICAL COMMUNICATION AND NETWORKING
2 Marks Questions and Answers
UNIT - 1 - INTRODUCTION
1. Define a fiber optic system.
Fiber optic system consists of a fiber optic cable, a light source and a light detector. The
optic fiber is used to carry the light beam from one place to another.
2. What are the uses of optical fibers?
a) To transmit the information of telephone communication, computer data, etc.
which are in the form of coded light signals
b) To transmit the optical images (Example : Endoscopy)
c) To act as a light source at the inaccessible places.
d) To act as sensors to do mechanical, electrical and magnetic measurements.
3. Differentiate between glass and plastic fiber cables.
Fiber optic cables are made from glass and fiber. Glass has the lowest loss but it is brittle.
Plastic is cheaper and more flexible but has high attenuation.
4. Mention the advantages of optical fiber communication.
1. Large information capacity 2. Long distance transmission
3. Small size and low weight 4. Electrical isolation
5. Immunity to crosstalk and EMI 6. Increased signal security
7. Enhanced safety 8. Ruggedness and flexibility
9. System reliability and easy maintenance 10. Low cost
5. Define reflection.
The law of reflection states that the angle at which the ray strikes the interface is exactly
equal to the angle that the reflected ray makes with the imaginary perpendicular normal.
6. Define refraction.
Refraction occurs when light ray passes from one medium to another i.e. the light ray
changes direction at the interface. The refraction (bending) takes place because light travels at
different speed in different mediums.
7. What is Snell’s law?
Snell’s law states how light ray reacts when it meets the interface of two mediums having
different refractive indices. Hence it is the relationship at the interface of two mediums and is
given by
n1sinΦ1=n2 sinΦ2.
where n1 is the refractive index of medium 1
n2 is the refractive index of medium 2
Φ1 is the angle of incidence
Φ2 is the angle of refraction
AMSEC/ECE Prepared By : Mr.A.Natarajan, ASP/ECE
, EC2402 – OPTICAL COMMUNICATION AND NETWORKS
8. What is total internal reflection?
When the incidence angle is increased beyond the critical angle, the light ray does not pass
through the interface into the other medium. This gives the effect of mirror existing at the interface
with no possibility of light escaping outside the medium. In this, the angle of reflection is equal to
the angle of incidence. This action is called the total internal reflection of the beam.
9. What are the conditions for total internal reflection?
a) Light should travel from denser medium to rarer medium.
b) The angle of incidence should be greater than the critical angle of the denser Medium.
10. What is internal reflection?
When the reflection of light is of a less optically dense material, it is called internal
reflection.
11. What is external reflection?
When light travelling in a certain medium is reflected off an optically denser material it is
referred to as external reflection.
12. What is meant by refractive index of a material?
The amount of refraction or bending that occurs at the interface of two materials of
different densities is expressed as refractive index of two materials. It is the ratio between the
speed of light in air and the speed of light in the material and is given by n = c /v.
13. What is critical angle of incidence?
The critical angle is the angle of incidence that causes the refracted light to travel along the
interface between two different mediums. It is also defined as the minimum angle of incidence at
which the ray strikes the interface of two media and causes an angle of refraction equal to 900.
Critical angle of incidence Φc= sin-1 (n2 / n1) where n1 is the refractive index of medium1
n2 is the refractive index of medium 2
14. Define acceptance angle. (Nov 14)
The maximum angle ‘Φmax’ with which a ray of light can enter through the entrance end of
the fiber and still be totally internally reflected is called acceptance angle of the fiber.
15. Define acceptance cone.
Rotating the acceptance angle ‘Φmax’ around the fiber axis, a cone shaped pattern is
obtained. It is called the acceptance cone of the fiber input. In other words, the acceptance cone is
the angle within which the light is accepted into the core and is able to travel along the fiber.
16. Write the expression for the refractive index in graded index fibers.
n(r) =n1[1-2∆(r/a)α]1/2for 0 ≤ r ≤ a
= n1(1-2∆)1/2≈ n1 (1-∆) = n2for r ≥ a
r →radial distance from fiber axis a→ core radius
n1→ refractive index at the core n2→refractive index at the cladding
α →shape of the index profile ∆→ index difference
AMSEC/ECE Prepared By : Mr.A.Natarajan, ASP/ECE
, EC2402 – OPTICAL COMMUNICATION AND NETWORKS
17. What is the necessity of cladding for an optical fiber?
a) To provide proper light guidance inside the core
b) To avoid leakage of light from the fiber
c) To provide mechanical strength for the fiber
d) To protect the core from scratches and other mechanical damages
e) To protect the core from absorbing surface contaminants
18. Define relative refractive index difference.
∆ = n12 – nn12 ≈ n1-n2 / n1
Thus relative refractive index difference is the ratio between the refractive index difference (of
core and cladding) and refractive index of core.
19. Define Numerical aperture of a step index fiber. (Nov 14)
Numerical aperture (N.A) of the fiber is the light collecting capability of the fiber and is the
measure of the amount of light rays that can be accepted by the fiber. It is equal to the sine of
acceptance angle.
N.A = sinΦmax = (n12-n22)1/2 =n1(2∆)1/2 where n1and n2are the refractive indices of core
and cladding respectively and ∆ is the index difference
20.Give the expression for numerical aperture in graded index fibers.
N.A.(r)=N.A.(0) [1-(r/a) α]1/2 for r≤ a
Where N.A(0) = axial numerical aperture = (n12-n22)1/2
a is core radius and
α is the refractive index profile.
21. What is an index profile?
The index profile of an optical fiber is a graphical representation of the magnitude of the
refractive index across the fiber.
22. Define Mode-field diameter.
The mode-field diameter (MFD) is the fundamental parameter of a single mode fiber. This
can be determined from the mode field distribution of the fundamental LP01 mode.
23. Why do we prefer step index single mode fiber for long distance communication? Or List
the advantages of single mode fibers. (Nov 14)
Step index single mode fiber has
a) low attenuation due to smaller core diameter
b) higher bandwidth and
c) very low dispersion.
24. What is tunnel effect?
The leaky modes are continuously radiating their power out of the core as they propagate
along the fiber. This power radiation out of the waveguide due to a quantum mechanical
phenomenon is known as the tunnel effect.
AMSEC/ECE Prepared By : Mr.A.Natarajan, ASP/ECE
OPTICAL COMMUNICATION AND NETWORKING
2 Marks Questions and Answers
UNIT - 1 - INTRODUCTION
1. Define a fiber optic system.
Fiber optic system consists of a fiber optic cable, a light source and a light detector. The
optic fiber is used to carry the light beam from one place to another.
2. What are the uses of optical fibers?
a) To transmit the information of telephone communication, computer data, etc.
which are in the form of coded light signals
b) To transmit the optical images (Example : Endoscopy)
c) To act as a light source at the inaccessible places.
d) To act as sensors to do mechanical, electrical and magnetic measurements.
3. Differentiate between glass and plastic fiber cables.
Fiber optic cables are made from glass and fiber. Glass has the lowest loss but it is brittle.
Plastic is cheaper and more flexible but has high attenuation.
4. Mention the advantages of optical fiber communication.
1. Large information capacity 2. Long distance transmission
3. Small size and low weight 4. Electrical isolation
5. Immunity to crosstalk and EMI 6. Increased signal security
7. Enhanced safety 8. Ruggedness and flexibility
9. System reliability and easy maintenance 10. Low cost
5. Define reflection.
The law of reflection states that the angle at which the ray strikes the interface is exactly
equal to the angle that the reflected ray makes with the imaginary perpendicular normal.
6. Define refraction.
Refraction occurs when light ray passes from one medium to another i.e. the light ray
changes direction at the interface. The refraction (bending) takes place because light travels at
different speed in different mediums.
7. What is Snell’s law?
Snell’s law states how light ray reacts when it meets the interface of two mediums having
different refractive indices. Hence it is the relationship at the interface of two mediums and is
given by
n1sinΦ1=n2 sinΦ2.
where n1 is the refractive index of medium 1
n2 is the refractive index of medium 2
Φ1 is the angle of incidence
Φ2 is the angle of refraction
AMSEC/ECE Prepared By : Mr.A.Natarajan, ASP/ECE
, EC2402 – OPTICAL COMMUNICATION AND NETWORKS
8. What is total internal reflection?
When the incidence angle is increased beyond the critical angle, the light ray does not pass
through the interface into the other medium. This gives the effect of mirror existing at the interface
with no possibility of light escaping outside the medium. In this, the angle of reflection is equal to
the angle of incidence. This action is called the total internal reflection of the beam.
9. What are the conditions for total internal reflection?
a) Light should travel from denser medium to rarer medium.
b) The angle of incidence should be greater than the critical angle of the denser Medium.
10. What is internal reflection?
When the reflection of light is of a less optically dense material, it is called internal
reflection.
11. What is external reflection?
When light travelling in a certain medium is reflected off an optically denser material it is
referred to as external reflection.
12. What is meant by refractive index of a material?
The amount of refraction or bending that occurs at the interface of two materials of
different densities is expressed as refractive index of two materials. It is the ratio between the
speed of light in air and the speed of light in the material and is given by n = c /v.
13. What is critical angle of incidence?
The critical angle is the angle of incidence that causes the refracted light to travel along the
interface between two different mediums. It is also defined as the minimum angle of incidence at
which the ray strikes the interface of two media and causes an angle of refraction equal to 900.
Critical angle of incidence Φc= sin-1 (n2 / n1) where n1 is the refractive index of medium1
n2 is the refractive index of medium 2
14. Define acceptance angle. (Nov 14)
The maximum angle ‘Φmax’ with which a ray of light can enter through the entrance end of
the fiber and still be totally internally reflected is called acceptance angle of the fiber.
15. Define acceptance cone.
Rotating the acceptance angle ‘Φmax’ around the fiber axis, a cone shaped pattern is
obtained. It is called the acceptance cone of the fiber input. In other words, the acceptance cone is
the angle within which the light is accepted into the core and is able to travel along the fiber.
16. Write the expression for the refractive index in graded index fibers.
n(r) =n1[1-2∆(r/a)α]1/2for 0 ≤ r ≤ a
= n1(1-2∆)1/2≈ n1 (1-∆) = n2for r ≥ a
r →radial distance from fiber axis a→ core radius
n1→ refractive index at the core n2→refractive index at the cladding
α →shape of the index profile ∆→ index difference
AMSEC/ECE Prepared By : Mr.A.Natarajan, ASP/ECE
, EC2402 – OPTICAL COMMUNICATION AND NETWORKS
17. What is the necessity of cladding for an optical fiber?
a) To provide proper light guidance inside the core
b) To avoid leakage of light from the fiber
c) To provide mechanical strength for the fiber
d) To protect the core from scratches and other mechanical damages
e) To protect the core from absorbing surface contaminants
18. Define relative refractive index difference.
∆ = n12 – nn12 ≈ n1-n2 / n1
Thus relative refractive index difference is the ratio between the refractive index difference (of
core and cladding) and refractive index of core.
19. Define Numerical aperture of a step index fiber. (Nov 14)
Numerical aperture (N.A) of the fiber is the light collecting capability of the fiber and is the
measure of the amount of light rays that can be accepted by the fiber. It is equal to the sine of
acceptance angle.
N.A = sinΦmax = (n12-n22)1/2 =n1(2∆)1/2 where n1and n2are the refractive indices of core
and cladding respectively and ∆ is the index difference
20.Give the expression for numerical aperture in graded index fibers.
N.A.(r)=N.A.(0) [1-(r/a) α]1/2 for r≤ a
Where N.A(0) = axial numerical aperture = (n12-n22)1/2
a is core radius and
α is the refractive index profile.
21. What is an index profile?
The index profile of an optical fiber is a graphical representation of the magnitude of the
refractive index across the fiber.
22. Define Mode-field diameter.
The mode-field diameter (MFD) is the fundamental parameter of a single mode fiber. This
can be determined from the mode field distribution of the fundamental LP01 mode.
23. Why do we prefer step index single mode fiber for long distance communication? Or List
the advantages of single mode fibers. (Nov 14)
Step index single mode fiber has
a) low attenuation due to smaller core diameter
b) higher bandwidth and
c) very low dispersion.
24. What is tunnel effect?
The leaky modes are continuously radiating their power out of the core as they propagate
along the fiber. This power radiation out of the waveguide due to a quantum mechanical
phenomenon is known as the tunnel effect.
AMSEC/ECE Prepared By : Mr.A.Natarajan, ASP/ECE
