-ADVANTAGES OF OPTICAL COMMUNICATION
 Low transmission loss.
 Wide bandwidth.
 Small size and weight.
Immunity to interference (EMI &
EMP).
 Electrical insulation (No arcing,
sparking, ground loops, hazards).
 Signal security (banking, computer
networks, military systems).
Abundant raw material
‫نخلى بالنا بقى من المقارنتين اللى جايين دول عشان مهمين جدا وياريت‬‫نراجع عليهم يوم السبت بليل عشان مننسهاش‬
 PROBLEMS
1-a) Using simple ray theory, describe the mechanism
for the transmission of light within an optical fiber.
b) Briefly discuss with the aid of a suitable diagram
what is meant by the acceptance angle for an optical
fiber.
c) Show how this is related to the fiber numerical
aperture and the refractive indices for the fiber core
and cladding.
d) An optical fiber has a numerical aperture of 0.20
and a cladding refractive index of 1.59. Determine:
(i) The acceptance angle for the fiber in water which
has a refractive index of 1.33;
(ii) The critical angle at the core–cladding interface.
Comment on any assumptions made about the fiber.
Solution
A)if the incident light on the interface by angle larger
than 𝑄𝑐 then the light could be still bounded in the
core till its end, while if Q<𝑄𝑐 light will be lost in
cladding
b)if Ө𝑜
Ө
𝑄𝑖𝑛 which will result in 𝑄𝑖𝑛 <𝑄𝑐
which will cause unbounded ray so we can say that
if Ө𝑜 >Ө𝑜 𝑚𝑎𝑥 we'll have unbounded ray
‫ االثبات ده‬mid-term revision‫راجع من ال‬c)
d)using this rule NA=√𝑛1 2 − 𝑛2 2 =sin Ө𝑜 𝑚𝑎𝑥
𝑄𝑐 =sin−1
𝑛2
𝑛1
i-62.8𝑜 , ii-56.27𝑜
2-The velocity of light in the core of a step index fiber
is 2.01 × 108 m s−1, and the critical angle at the core–
cladding interface is 80°. Determine the numerical
aperture and the acceptance angle for the fiber in air,
assuming it has a core diameter suitable for
consideration by ray analysis. The velocity of light in a
vacuum is 2.998 × 108 m s−1.
Solution
Using the rules NA=√𝑛1 2 − 𝑛2 2 =sin Ө𝑜 𝑚𝑎𝑥
𝑄𝑐 =sin−1
V=
𝑐
𝑛2
𝑛1
𝑛1
Then NA=.238
Ө𝑜 𝑚𝑎𝑥 =13.787 degree.
3-a) Define the relative refractive index difference for
an optical fiber.
b) Show how it may be related to the numerical
aperture.
c) A step index fiber with a large core diameter
compared with the wavelength of the transmitted
light has an acceptance angle in air of 22° and a
relative refractive index difference of 3%. Estimate the
numerical aperture and the critical angle at the core–
cladding interface for the fiber.
Solution
a,b)so easy
c)Ө𝑜 𝑚𝑎𝑥 =22𝑜 then
NA=sin−1 Ө𝑜 𝑚𝑎𝑥 =.3746=√𝑛1 2 − 𝑛2 2
But 𝑛1 − 𝑛2 = .03 then𝑛1 = 𝑛2 + .03
‫𝑛 ومنها اجيب‬2 ‫𝑛 تحت الجذر اقدر اجيب قيمة‬1 ‫وبالتعويض عن قيمة‬
𝑛
‫= 𝑐𝑄 نالقى ان‬sin−1 2 ‫𝑛 وبعدين من القانون ده‬1 ‫قيمة‬
𝑛1
𝑄𝑐 = 80.84𝑜
‫ملحوظة بس على المسألة دى‬
𝑛1 − 𝑛2 = relative refractive index difference-1
𝑛1 −𝑛2
𝑛1
= fractional refractive index difference-2
5-Describe with the aid of simple ray diagrams:
(a) the multimode step index fiber;
(b) the single-mode step index fiber.
Compare the advantages and disadvantages of these
two types of fiber for use as an optical channel
Solution
‫وكمان معاها المقارنة اللى فى اول المراجعة ‪.‬‬