14. « Refraction is the bending of
light as it passes from one
substance to another »
15. « Refraction is the bending of
light as it passes from one
substance to another »
Air
Glass
16. ➢ Vocabulary of Refraction
incident ray
Air θinc
interface
Water
17. ➢ Vocabulary of Refraction
incident ray
Air θinc
interface
θrefr
Water
refracted ray
18. ➢ Vocabulary of Refraction
incident ray
reflected
ray
Air θinc
θrefl
interface
θrefr
Water
refracted ray
19. ➢ Vocabulary of Refraction
θinc= θrefl
incident ray
reflected
ray
Air θinc
θrefl
interface
θrefr
Water
refracted ray
20. ➢ Vocabulary of Refraction
θinc= θrefl
incident ray
reflected
ray
Air θinc
θrefl
interface
θrefr
Water
refracted ray
21. ➢ Vocabulary of Refraction
θinc= θrefl
inc id e nt ra y
reflected
ray
Air θinc
θrefl
interface
θrefr
Water
P refracted ray
22. ➢ Vocabulary of Refraction
θinc= θrefl
incident ray
reflected
ray
Air θinc
θrefl
nair
interface
θrefr
Water
nwater
P refracted ray
23. ➢ Snell-Descartes Law
incident ray
Air θinc
nair
interface
θrefr
Water
nwater
refracted ray
24. ➢ Snell-Descartes Law
nair. sin θinc = nwater. sin θrefr
incident ray
Air θinc
nair
interface
θrefr
Water
nwater
refracted ray
25. ➢ Snell-Descartes Law
nair. sin θinc = nwater. sin θrefr
incident ray
Air θinc
nair
interface
θrefr
Water
nwater
refracted ray
n1. sin θinc = n2. sin θrefr
26. ➢ Snell-Descartes Law
nair =1
nair. sin θinc = nwater. sin θrefr
nwater =1.5
incident ray
θinc= 45°
Air θinc
nair
interface
θrefr
Water
nwater
refracted ray
n1. sin θinc = n2. sin θrefr
27. ➢ Snell-Descartes Law
nair =1
nair. sin θinc = nwater. sin θrefr
nwater =1.5
incident ray
θinc= 45°
Air θinc
nair
interface sin θrefr= (nair. sin θinc) / nwater
θrefr
Water
nwater
refracted ray
n1. sin θinc = n2. sin θrefr
28. ➢ Snell-Descartes Law
nair =1
nair. sin θinc = nwater. sin θrefr
nwater =1.5
incident ray
θinc= 45°
Air θinc
nair
interface
θrefr sin θrefr= (nair. sin θinc) / nwater
Water
nwater = 0.71/1.5
refracted ray = 0.47
n1. sin θinc = n2. sin θrefr
29. ➢ Snell-Descartes Law
nair =1
nair. sin θinc = nwater. sin θrefr
nwater =1.5
incident ray
θinc= 45°
Air θinc
nair
interface
θrefr sin θrefr= (nair. sin θinc) / nwater
Water
nwater = 0.71/1.5
refracted ray = 0.47
n1. sin θinc = n2. sin θrefr θrefr = arcsin 0.47
=28.13°
40. ➢ Optical Fiber
An optical fiber is a thread in glass or plastic which has the property to drive the light. To drive the light without leak or loss,
we have to avoid phenomenoms of refraction inside the thread. In order to avoid it, the thread is constituted by a
core (index n1) and a duct (index n2).
n2 Duct
Core
n1 θ1
θair
1. Write the Snell-Descartes law for the interface between air and the core of the optical fiber.
2. What will be the value of θ1 for θair = 25° , nair=1 and n1=1.4 ?
How will θ1 evolve if n1> 1.4 ?
41. ➢ Optical Fiber
An optical fiber is a thread in glass or plastic which has the property to drive the light. To drive the light without leak or loss,
we have to avoid phenomenoms of refraction inside the thread. In order to avoid it, the thread is constituted by a
core (index n1) and a duct (index n2).
n2 Duct
Core
n1 θ1
θair
n2
3. Find a condition on n2 to avoid the refraction and have a total reflection.
Clue : What is the critical value of an angle of refraction ?
n1 θ1
42. ➢ Cold or warm mirages ?
The index of air is decreasing with the temperature (n warm < ncold). Mirages are manifestations of this variation.
When the variation of temperature is important the refraction bends the path of the light, like with the pencil in
water, except here there is a succession of different interfaces.
n1
n2
n3
n4
n5
Actually, there are two kinds of mirages : warm and cold. Eye
Warm mirages can be observed in warm area, and cold mirages in cold
area.
Warm
1. Can you explain why warm mirages are in warm area ? Mirage
2. Same question for the cold mirage
Eye Cold
Mirage
43. Push on « Start »
button !
Push again when
you are ready to
have the correction !
44. ➢ Optical Fiber
An optical fiber is a thread in glass or plastic which has the property to drive the light. To drive the light without leak or loss,
we have to avoid phenomenoms of refraction inside the thread. In order to avoid it, the thread is constituted by a
core (index n1) and a duct (index n2).
n2 Duct
Core
n1 θ1
θair
1. Write the Snell-Descartes law for the interface between air and the core of the optical fiber.
nair. sin θ air = n1. sin θ 1
2. What will be the value of θ1 for θair = 25° , nair=1 and n1=1.4 ?
How will θ1 evolve if n1> 1.4 ?
nair. sin θ air = n1. sin θ 1 θ 1 = arcsin (nair. sin θ air /n1 )
= 17.6°
If n1 > 1.4, θ 1 < 17.6°
45. ➢ Optical Fiber
An optical fiber is a thread in glass or plastic which has the property to drive the light. To drive the light without leak or loss,
we have to avoid phenomenoms of refraction inside the thread. In order to avoid it, the thread is constituted by a
core (index n1) and a duct (index n2).
n2 Duct
Core
n1 θ1
θair
n2 ω2
3. Find a condition on n2 to avoid the refraction and have a total reflection.
Clue : What is the critical value of an angle of refraction ? ω1
n1 θ1
ω1= 180 – (90 + θ1 ) ω2 = 90° So, sin ω2 = 1
Snell-Descartes law : n1.sin ω1 = n2
If n1.sin ω1 > n2 , refraction can't occure.
For θ1 = 17.6° and n1 = 1.4
ω1 = 72.4° n2 <1.33
46. ➢ Cold or warm mirages ?
The index of air is decreasing with the temperature (n warm < ncold). Mirages are manifestations of this variation.
When the variation of temperature is important the refraction bends the path of the light, like with the pencil in
water, except here there is a succession of different interfaces.
n1
n2
n3
n4
n5
Actually, there are two kinds of mirages : warm and cold. Eye
Warm mirages can be observed in warm areas, and cold mirages in cold
areas.
Warm
1. Can you explain why warm mirages are in warm areas ? Mirage
In warm areas, air is warmer near of the ground. So n is increasing with
the height. The Snell-Descartes law attest to this bending.
n1>n2 θ1<θ2
n1<n2 θ1>θ2
2. Same question for the cold mirage.
Eye Cold
It's exactly the same phenomenon but Mirage
this time, n is decreasing with the height
