Chapter three
Light
Light is a form of energy.
The sun is the main source of energy on earth surface.
The energy from the sun is almost divided between heat and light.
The plants need light to carry out photosynthesis process to make food.
Light has a wave nature as it can reflect, refract, interfere and diffract.
Light is an electromagnetic wave which doesn't need medium to transfer through, it
consists of vibrating electric and magnetic fields, oscillate at equal frequency at the same
phase normal to each other and normal to the direction of propagation.
Light is a transverse wave consisting of crests and trough.
They travel at constant speed in space (3x𝟏𝟎𝟖 m/s).
The electromagnetic spectrum
When the light ray falls on a separating surface between two optical media which are
different in optical densities, part of it will reflect and the other part will refract neglecting
the absorbed part by medium.
1
Reflection of light
Reflection of light: it's the rebounding (return back) of the light rays in the same medium
when they meet a reflecting surface.
Laws of reflection
The angle of incidence = the angle of reflection.
The incident ray, the reflected ray and the normal to
the reflecting surface at the point of incidence all lie
in one plane normal to the reflecting surface.
When a light ray falls normally to the reflecting
surface, it reflects on itself.
Because the angle of incidence = the angle of
reflection = 0.
Refraction of light
Refraction of light: it's the bending (changing in the path) of the light rays when
passingbetween two mediumhave different optical density.
Occurrence of light refraction isDue to the difference in the speed of light in the two
media.
Optical density: the ability of the medium to bend light rays.
2
The ratio between the sine of the angle of
incidence in the 1st medium to the sine of angle of
refraction in the 2nd medium is equal to the ratio
between the speed of light in the two media.
𝑉1 sin 𝛷 𝜆1
𝑛2
=
=
= 1𝑛2 =
𝑉2 sin 𝜃 𝜆2
𝑛1
The incident ray, the refracted ray and the normal
to the separating surface (interface) at the point
of incidence all lie in one plane to the separating
surface.
Relative refractive index:It's the ratio between the speed of light in the 1st medium to the
speed of light in the 2nd medium.
It's the ratio between the absolute refraction index of the second medium to the absolute
refraction index of the first one.
It's the ratio between the sin the angle of incidence in the first medium to sin the angle of
refraction in the second medium.
Absolute refractive index for a medium: It's the ratio between the speed of light in space
(or air) to the speed of light in this medium.
Note:
The speed of light in space (c) is one of the physical constants in the universe. (c = 3 x 108
m/s).
The speed of light in space is greater than the speed of light in any other media.
𝑐
𝑐
𝑐
𝑐 𝑉1 sin 𝛷
𝑛1 =
, 𝑆𝑜 𝑉1 =
& 𝑛2 =
, 𝑆𝑜 𝑉2 =
&
=
=
𝑉1
𝑛1
𝑉2
𝑛1 𝑉2 sin 𝜃
𝑐
𝑛1
𝑐
𝑛2
sin 𝛷 𝑛2
=
= 1𝑛2, 𝑆𝑜 𝑛1 ∗ 𝑠𝑖𝑛𝛷 = 𝑛2 ∗ 𝑠𝑖𝑛𝜃 𝑆𝑛𝑒𝑙𝑙 ′ 𝑠 𝐿𝑎𝑤
sin 𝜃 𝑛1
Snell's Law: the product of the absolute refractive index of the first media and sine the
angle of incidence in it equal the product of the absolute refractive index of the second
medium and sine the refraction angle in it.
We can use refraction to analyze a white light ray into its components of different
wavelengths (light spectrum-ROYG.BIV) as the absolute refractive index of the medium
differs according to the wavelength of each color.
3
When a light ray passes from a medium to
denser medium, it bends towards the normal
(𝛷 > 𝜃, 𝑉1 > 𝑉2, 𝑛1 < 𝑛2, 1𝑛2 > 1)
When a light ray passes from a medium to
less denser medium, it bends away from the
normal (𝛷 > 𝜃, 𝑉1 > 𝑉2, 𝑛1 < 𝑛2, 1𝑛2 > 1)
When a light ray falls normal on the
separating surface, it doesn't refract
Interference of light
Interference of light: it’s superposition of two light waves having same frequency,
amplitude and in phase (same velocity & direction) so the intensity of the light reinforced at
some positions (illuminated or bright fringes) and it’s weakened in other position (dark
fringes).
Double Slit experiment (Thomas Young):
Monochromatic light: it’s light source with constant wavelength (red light recommended).
Like sodium lamp.
Coherent sources: they are light sources that emit waves of same frequency, amplitude and
in phase (same direction & velocity).
Young’s double slits experiment:
Usage:
4
Used to explain the interference of light waves and calculate the wave length of any
monochromatic light.
Steps:
 A source of monochromatic light is placed at a suitable distance from a rectangular
slit (a) at screen S1, a cylindrical waves pass towardanother screen S2 having two
narrow slits (b,c).
 Cylindrical waves (from the two slits acting as two coherent sources) interfere with
each other.
 The result interference is received on another screen in form of bright and dark
fringes.
 We can calculate the wavelength of the monochromatic light from the following
equation:
𝛥𝑦 =
λ∗R
Δy ∗ d
𝑠𝑜 λ =
𝑑
R
Where:
λ: is the wavelength of the monochromatic light.
R: is the distance between the two double slits screen and the receiving one.
d: is the distance between the two slits.
Notes:
If the different path = m*λ where m is (0, 1, 2,…)bright fringe appears.
If the different path = (m+0.5)*λ where m is (0, 1, 2,…)dark fringe appears.
The central fringe is always bright due to the path difference is zero so the constructive
interference occurs (two crests or two troughs meeting). The intensity of light increase
forming bright fringe.
The condition to obtain clear interference fringes:
 Present of two coherent light sources.
 The coherent sources must be close to each other (decrease d).
 Used monochromatic light source with high wavelength (increase λ).
 The size of any of the slits must be smaller than the wavelength of the incident
waves. (Diffraction).
5
Diffraction of light
Diffraction
Airy Disk
Diffraction of light: it’s the formation of a circular spot with bright and dark fringes (called
Airy disk) due to flaring out monochromatic light as it passes through a small hole or solid
edge (instead of passing in straight line).
Critical Angle and Total internal reflection
Critical angle (Фc): it’s the angle of incidence in the denser medium when the angle of
refraction in the less one is 90 ˚.
From the Snell’s law
n1*sinФ = n2*sinθ As θ=90˚ For Фc so sin90=1.
𝑛2
n1*sinФc = n2 so sinФc = =1n2.
𝑛1
1
For air sinФc= .
𝑛1
Notes:
n1 is always greater than n2.
Increasing the difference between n1 and n2 decreasing the critical angle.
The absolute refractive index of a medium = the reciprocal of the sin of critical angle.
6
The value of the critical angle is inversely proportional to the absolute refractive index of
the medium.
Application of total internal reflection
Optical fibers:
Optical fiber:it’s a thin flexible tube made of a transparent material (like glass fibers) used
to transmitting light energy based on total internal reflection as following:
In medical (Endoscopes): transmitting images of internal parts of body.
In communication: using laser to transmit the electric signals.
How it works: Light rays enter the optical fiber with angle of incidence greater than the
critical angle so they suffer total internal reflection till it emerges from the other end.
Totally reflecting prisms:
7
Triangle glass prism of (90ᵒ-45ᵒ-45ᵒ) used in:
Changing the path of light rays by 90ᵒ or 180ᵒ (angle of emergence).
Periscopes.
Binocular.
8
Changing the upper and lower light rays of an image with each other.
Notes:
Totally reflecting prisms is better than plane mirrors in reflecting metallic surface because
they reflect 100% of incident light rays and don’t lose their luster but mirrors do.
The face of reflecting prism covered with a thin layer of Cryolite (Aluminum fluoride) or
magnesium fluoride because they have a less refractive index than glass to minimize the
loss of light intensity.
Mirage:
Mirage: it’s a natural phenomenon occurs in hot regions at noon in hot day where false
images appear.
Examples: See false images above water
surface, see water image on the paved road.
Explanation: the light rays pass from cold air
layers to hotter one so refract away from the
normal till the angle of incidence is greater
than the critical angle so total reflection occurs
producing false image.
Deviation in triangular prism
Angle of deviation (α): it’s the angle between the extension of the incidence ray and the
emergent ray.
Prim rules Prove:
Ф1: the angle of incidence (from air to glass).
Ѳ1: the angle of refraction.
Ф2: the internal angle of incidence (2nd).
Ѳ2: angle of deviation.
A: the prism angle, Apex, vertex angle or
reflective angle.
The rule of prism angle
A + E = 180ᵒ (OBEC cyclic quadrilateral).
Ѳ1 + Ф2 + E = 180ᵒ (BCE triangle).
9
A = Ѳ1 + Ф2
The rules of angle of deviation:
α = 1 + 2 (exterior angle).
Ф1= 1 + Ѳ1. So 1 = Ф1-Ѳ1
Ѳ2 = 2 + Ф2. So 1 = Ѳ2-Ф2
α = (Ф1-Ѳ1) + (Ѳ2-Ф2) = (Ф1+Ѳ2) - (Ѳ1+Ф2)
α = (Ф1+ Ѳ2) – A
The angle of deviation (α)&the angle of incidence (Ф) relation.
From experiment: Increasing angle of incidence decreasing
angle of deviation till a certain value (Minimum deviation)
then increase again.
Minimum deviation position view:
The angle of incidence (Ф1) = the angle of emergency (Ѳ2).
The internal angle of incidence (Ф2) = the angle of
refraction (Ѳ1).
The white light separated into its seven colors.
𝑛=
sin Ф1 sin Ѳ2
=
sin Ѳ1 sin Ф2
Rules of prism angle at min deviation:
A = Ф2+Ѳ1 = 2Ѳₒ so Ѳₒ =
𝐴
2
αₒ = Ф1+Ѳ2-A = 2Фₒ-2Ѳₒ, so Фₒ =
𝛼ₒ+2Ѳ
2
=
𝛼ₒ+𝐴
2
𝜶ₒ+𝑨
𝐬𝐢𝐧(
)
sin Фₒ
𝟐
𝑛=
, 𝑆𝑜 𝒏 =
𝑨
sin Ѳₒ
𝐬𝐢𝐧( )
𝟐
Where:
n: the refractive of index of prism.
αᵒ: the angle of min deviation.
A: the prism angle.
10
Dispersion of light by a triangle prism:
When a beam of white light falls on one face of refraction side of triangle prism adjusted at
minimum deviation position it dispersed into seven colors having different wavelengths
called light spectrum.
At min deviation 𝑛 =
𝛼ₒ+𝐴
)
2
𝐴
sin( )
2
sin(
The prism angle (A) is constant
The angle of minimum deviation (αₒ) is directly proportional to the refractive index (n).
The refractive index (n) is inversely proportional to the velocity of light in the medium (𝑛 =
𝐶
𝑉
).
The velocity of the light in the media is directly proportional to the wavelength (λ) (𝑉 = 𝜆 ∗
𝜐).
The refractive index of light (n) is inversely proportional to the wavelength (λ)
So the angle of deviation (α) is inversely proportional to the wavelength (λ).
The red color has max wavelength (λ) so has min deviation (α), but the violet color has min
wavelength (λ) so has max deviation (α).
Thin prism
Thin prism: it’s a prism whose apex angle is very small about 10⁰.
Thin prism is always in a position of min deviation.
In case of small angles, the angle in radians = it’s sine in
degree.
𝑛=
sin(
𝛼ₒ+𝐴
2
𝐴
sin( )
2
So
𝛼ₒ
𝐴
)
𝛼ₒ+𝐴
=
2
𝐴
2
=
𝛼ₒ + 𝐴 𝛼ₒ
= +1
𝐴
𝐴
= 𝑛 − 1, 𝜶ₒ = 𝑨(𝒏 − 𝟏)
𝛼ₒ𝑏 = 𝐴(𝑛𝑏 − 1)𝑓𝑜𝑟 𝑏𝑙𝑢𝑒 𝑐𝑜𝑙𝑜𝑟,
𝛼ₒ𝑟 = 𝐴(𝑛𝑟 − 1)𝑓𝑜𝑟 𝑟𝑒𝑑 𝑐𝑜𝑙𝑜𝑟.
Where:
𝛼ₒ𝑏, 𝛼ₒ𝑟 is the angle of deviation of the blue and red color respectively.
𝑛𝑏, 𝑛𝑟 is the refractive index of the glass to the blue and red color respectively.
11
The angular size between red and blue rays in the thin prism:
It’s the angle included between the red and the blue light rays Or it’s the difference
between the angle of deviation f blue and red light rays.
The mathematic equation is 𝛂ₒ𝐛 − 𝛂ₒ𝐫 = 𝐀(𝐧𝐛 − 𝐧𝐫).
The dispersive power (ω): it’s the ratio between the angular size between two colors and
their average deviation.
The yellow lies between the red and blue lights so the average deviation between red and
blue colors is that of the yellow color.
𝛼ₒ𝑦 = 𝐴(𝑛𝑦 − 1), 𝑛𝑦 =
𝑆𝑜 𝝎 =
𝑛𝑏 + 𝑛𝑟
2
𝜶ₒ𝒃 − 𝜶ₒ𝒓 𝒏𝒃 − 𝒏𝒓
=
𝜶ₒ𝒚
𝒏𝒚 − 𝟏
Notes:
The glass block doesn’t disperse the white light into seven colors or deviate its path
because we can consider it as two identical adjacent glass prisms place opposite to each
other; the action caused by one of them (dispersion or deviation) is cancelled by the other.
From the symmetry of (α, φ) curve, the light ray falls on the face of the glass prism has two
values of angle of incidence for every angle deviation except the min deviation.
The angle of incidence corresponding to the min deviation value = the average of the
previous two values.
12
Experiment
Tracing the path of ray through a triangle glass
prism and verifying the laws of refraction
through:
Devices: Equilateral glass prism (A=60̊) – pins –
protractor – ruler
Steps:
1- Place the prism on one of its bases on sheet of paper and mark its position using
pencil.
2- Fix 2 pins (a and b) vertically so that the line joining them represents the incident ray.
3- Look at the opposite face of the prism to see the images of the 2 pins and fix other 2
pins (c and d) along their extensions so that they represents ray of emergence.
4- Remove the prism and the pins and draw the lines (ab, bc, cd) (air – glass – air ).
5- Draw the extensions ab and cd and measure the acute angle between them α.
6- Measure the angles 1, 2, 3, 4 which are φ1, ϴ1, φ2, ϴ2.
7- Repeat the previous steps several times changing the angle of incidence and tabulate
the results.
Trial
prism
angle
angle of
incidence
angle of
refraction
internal
angle of
incidence
angle of
emergence
angle of
deviation
A
φ1
ϴ1
φ2
ϴ2
α
1
…………. …………….
……………………
………………….. …………………
………………
2
…………. …………….
……………………
………………….. …………………
………………
3
…………. …………….
……………………
………………….. …………………
………………
Conclusion:
From the experiment we find that:
A=ϴ1+φ2, α=φ1+ϴ2-A
13
Index of Refraction
Material
Index
Vacuum
1.00000
Air at STP
1.00029
Ice
1.31
Water at 20 C
1.33
Acetone
1.36
Ethyl alcohol
1.36
Sugar solution(30%)
1.38
Fluorite
1.433
Fused quartz
1.46
Glycerine
1.473
Sugar solution (80%)
1.49
Typical crown glass
1.52
Crown glasses
1.52-1.62
Spectacle crown, C-1
1.523
Sodium chloride
1.54
Polystyrene
Carbon disulfide
Flint glasses
Heavy flint glass
Extra dense flint, EDF-3
1.55-1.59
1.63
1.57-1.75
1.65
1.7200
Methylene iodide
1.74
Sapphire
1.77
Rare earth flint
1.7-1.84
Lanthanum flint
1.82-1.98
Arsenic trisulfide glass
2.04
Diamond
2.417
14
Take Away
Define:
Electromagnetic waves, reflection of light, refraction of light, angle of incidence, angle of
refraction, angle of reflection, laws of light reflection, relative index of refraction, absolute
index of refraction, Snell’s law, Interference of light, Diffraction of light, critical angle, total
internal reflection, Angle of deviation, thin prism, the dispersive power,
Compare:
1- Refraction and diffraction
Give reason:
1- The light ray falling normally to the reflecting surface rebounding on itself.
2- The absolute refractive index of any medium is always greater than one.
3- The central fringe in the Young experiment is always bright.
4- The totally reflecting prism is preferred than the reflecting mirror.
5- The optical fiber is used in seeing the internal parts of the human.
6- The thin prism used to disperse the white light into seven spectrum colors.
7- The angle of deviation of the red color is less than the violet color one.
What’s meant by:
1- The relative refraction index between glass and water = 0.875.
2- The absolute refractive index of diamond = 2.4.
Mention the factors affected the following:
1- Mirage.
2- Obtaining clear interference fringes in Young’s experiment.
Prove the following relation:
1- A=ϴ1+φ2 (with drawing).
2- α=φ1+ϴ2-A (mention the angles names).
𝛼ₒ+𝐴
2
𝐴
sin
2
sin
3- n=
(mention when this equation is used).
4- αₒ=A*(n-1) (mention when this equation is used).
5- 𝑛1 ∗ sin φ = 𝑛2 ∗ sin 𝛳
15
Mention the mathematic equation, slope value of C if it’s required for the following:
Problems:
1- A light ray falls on the surface of glass of absolute refractive index 1.5 with angle of
incidence = 30⁰. Find the angle of refraction.
2- Given that the absolute index of refraction of water and glass are 4/3, 3/2
respectively; find the relative index of refraction from water to glass and from glass
to water.
3- If you know refraction index of glass = 1.6 and water = 1.33; calculate the critical
angle for the glass, water and from glass to water.
4- A ray of light falls normal to one face of a glass prism of angle 45⁰ and emerges
tangent to the opposite face of the prism find:
- Refractive index of the prism
- The angle of deviation
- Velocity of the light in the glass
5- An equilateral glass prism of index 1.732, if its smallest angle of deviation when it’s
totally immersed in liquid is 32⁰; find the refractive index of the liquid.
6- Trace the path of incident light ray (showing the angles) till it emerges from the
prism in the following figure (n glass = 1.5).
16
7- A small lamp is put at depth 4 cm from the surface of water; if the area of the
smallest disk to make the lamb light unseen is 28.3 𝑐𝑚2 . Calculate the refractive
index of water.
8- A thin prism of angle 8⁰ made of crown glass if the refraction index of the crown
glass to the red light is 1.514 and blue light is 1.523; find
- The angular size.
- Dispersive power.
9- Calculate the smaller angle of incidence of light ray falls at one face of triangle glass
prism of vertex 75⁰; it emerges tangent to the other side knowing that ng=√2.
10If you know that the min deviation angle for a Normal triangle prism is 48.5⁰ at
angle of incidence = 37⁰ calculate.
- The emergence angle of the light ray.
- Angle of the prism (Apex).
- The refractive index of the prism material.
11A ray of light traveling in air heads the midpoint of one side of sheet with
refractive index 1.48 as shown the No of reflection process inside the sheet.
12The following table gives the relation between the angle of incidence φ in air
and the corresponding angle of refraction ϴ in solid transparent medium.
φ
0
20
30
40
50
60
70
80
90
ϴ
x
13.2
19.5
25.4
30.7
35.3
38.8
41
y
Plot a graphical relation between (sin φ) on the ordinate and the corresponding (sin
ϴ) on the abscissa; from the graph find
- The value of each x and y.
- The absolute refractive index of considered medium.
The critical angle of considered medium
17
Model answer
Define:
Electromagnetic wave: it's the result of oscillating electric field and magnetic field with
same frequency normal to each other and normal to the direction of wave motion.
Reflection of light: it's the rebounding of the light ray when meeting a reflected surface like
mirror.
Refraction of light: it’s the changing in the light ray direction when passing from a
transparent medium to another one having different optical density.
Angle of incidence: it's the angle between the incoming ray (incident ray) and the normal to
the reflected or refracted surface.
Angle of reflection: it's the angle between the reflected ray and the normal to the reflected
surface.
Angle of refraction: it’s the angle between the refracted ray and the normal to the
refracted surface.
Laws of reflection: the angle of incidence equal the angle of reflection, the incoming ray
(incident ray), the reflected ray and the normal to the reflected surface at the point of
incidence all place in the same plane normal to the reflected surface.
Relative index of refraction: it’s the ratio between the velocity in the first medium to the
velocity in the other one; Or the ratio between the absolute index of refraction in the
second medium to the absolute index of refraction in the first one; Or the ration between
the angle of incidence in the first medium to the angle of refraction in the second one; Or
the ration between the wavelength in the first medium to it in the second one.
Absolute refractive index: it’s the ratio between the velocity in air or vacuum to it in the
media and it always greater than one; Or the reciprocal of the sine of the critical angle of
the medium when the light travel from this media to the air or vacuum.
Snell's Law: the absolute refractive index of the first medium times the sine the angle of
incidence on it equal to the absolute refractive index of the second medium times the sine
the angle of refraction on it.
Interference in light: it's the superposition of two light waves same in wavelength,
amplitude and in phase as the light intensity reinforced in some region (bright fringes) and
vanished in others (dark fringes).
18
Diffraction of light: it's the formation of a circular spot with bright and dark fringes due to
the flaring out monochromatic light as it passes through a small hole or solid edge instead
of passing in straight line.
Critical angle: it's the angle of incidence in the higher dense medium when the angle of
refraction in the lower one equal 90̊.
Total internal reflection:when the angle of incidence in the higher dense media is greater
than the critical angle between the two media the light ray will not refract into the less
dense medium it will suffer total internal reflection in the dense one.
Angle of deviation: it's the angle between the extension of the incoming ray (incident ray)
and the emergent ray in the prism.
Thin prism: it's a prism with very small Apex angle about 10̊ and the angle of deviation is
always at minimum value (minimum deviation position).
The dispersive power: it’s the ratio between the angular size between two colors and their
average deviation angle.
Give reason:
1- Because the angle of incidence = zero and according to the laws of reflection angle f
incidence = angle of reflection = zero so the reflected ray will be normal to the
reflected surface (as the incidence ray) but in the opposite direction of the incidence
ray.
2- Because the absolute refractive index is the ratio between the light speed in the air
or vacuum to it in the media and the speed of light in air or vacuum is the greatest
one.
3- Because at the center the path difference between the two waves = zero so
constructive interference occurs according to this condition (Path Diff=m*λ m= (0,1,
2,…)) so the bright fringe occurs in the center.
4- Because light suffer total internal reflection in prism (100%) but in mirror no
(efficiency is less than 100%), also the luster of the mirrorloses by time decreasing its
reflected ability but prism no.
5- Because it's a thin glass tube designed as the angle of incidence of light ray on it is
always greater than the critical angle so the light ray suffer total internal reflection
it's used to transfer image.
6- Because the white light consists of seven colors differ in wavelength and according to
the refraction laws
sin 𝛷
sin 𝜃
=
𝜆1
so differ wavelength has differ angle or refraction.
𝜆2
19
7- Because the angle of deviation is inversely proportional to the wavelength 𝛼 =
𝑐
𝐴(𝑛 − 1), 𝑛 = , 𝑉 = 𝑣 ∗ 𝜆) and the red color has a wavelength greater than the
𝑉
blue color, so the angle of deviation of the red color is less than the blue one.
What's meant by
1- The ratio between the sine of angle of incidence in glass to the sine of the angle of
refraction in the water = 0.875 or the ratio between the speed of light in glass to the
speed of light in the water = 0.875 or the ratio between the absolute refractive index
in the water to the absolute refractive index in glass = 0.875𝑔𝑛𝑤 =
sin 𝛷
sin 𝜃
=
𝑉𝑔
𝑉𝑤
=
𝑛𝑤
𝑛𝑔
2- The ratio between the light speed in the air to the light speed in the diamond = 2.4
Mention the factors:
1- Mirage occurs under two condition hot weather and the angle of incidence is greater
than the critical angle.
2- According to the rule (∆𝑦 =
𝜆∗𝑅
𝑑
)
So the factor is
1- Type of monochromatic light (according to the wavelength changing)
2- The distance between the two slits (d)
3- The distance between the two screen (slits screen and the receiving one)
4- The size of each slit comparing to the light wavelength.
Mention the equation and slope
1- 𝑠𝑖𝑛𝛷 = 1𝑛2 ∗ 𝑠𝑖𝑛𝜃, 𝑠𝑙𝑜𝑝𝑒 = 1𝑛2
2- 𝜆 =
∆𝑦
𝑅
∗ 𝑑, 𝑠𝑙𝑜𝑝𝑒 =
∆𝑦
𝑅
3- 𝛼˳ = 𝐴𝑛 − 𝐴, 𝑆𝑙𝑜𝑝𝑒 = 𝐴, 𝐶 = −𝐴.
4- 𝛼˳ = 𝐴(𝑛 − 1), 𝑆𝑙𝑜𝑝𝑒 = 𝑛 − 1.
Problems
1- Given: 𝑛𝑔 = 1.5, 𝜃 = 30
Request: Φ
Soln: 𝑛1 ∗ 𝑠𝑖𝑛𝛷 = 𝑛2 ∗ 𝑠𝑖𝑛𝜃 = 1 ∗ 𝑠𝑖𝑛30 = 1.5 ∗ 𝑠𝑖𝑛𝜃, 𝑆𝑜 𝜃 = 19˚ 28ˊ
4
3
3
2
2- Given: 𝑛𝑤 = , 𝑛𝑔 =
20
Request: 𝑤𝑛𝑔, 𝑔𝑛𝑤
Soln: 𝑤𝑛𝑔 =
𝑛𝑔
𝑛𝑤
3
2
4
3
=
9
𝑛𝑤
8
𝑛𝑔
= = 1.125, 𝑔𝑛𝑤 =
4
3
3
2
=
8
1
9
𝑤𝑛𝑔
= =
3- Given: 𝑛𝑔 = 1.6, 𝑛𝑤 = 1.33
Request: 𝛷𝑐𝑔, 𝛷𝑐𝑤, 𝛷𝑐𝑔𝑤
Soln: 𝑛1 ∗ 𝑠𝑖𝑛𝛷𝑐 = 𝑛2, 𝑠𝑖𝑛𝛷𝑐 =
1
1.33
, 𝛷𝑐𝑤 = 48˚ 45ˊ, 𝑠𝑖𝑛𝛷𝑐𝑔𝑤 =
𝑛2
, 𝑠𝑖𝑛𝛷𝑐𝑔 =
𝑛1
1.33
1.6
1
1.6
, 𝛷𝑐𝑔 = 38˚ 40ˊ, 𝑠𝑖𝑛𝛷𝑐𝑤 =
, 𝛷𝑐𝑔𝑤 = 56˚13ˊ
4- Given: 𝛷1 = 𝑍𝑒𝑟𝑜, 𝐴 = 45˚, 𝜃2 = 90˚
Request: 𝑛𝑝, 𝛼, 𝑉𝑔
Soln: 𝐴 = 𝜃1 + 𝛷2, 𝛷2 = 45˚, 𝑛𝑔 ∗ 𝑠𝑖𝑛𝛷2 = 1, 𝑛𝑔 = 1.414, 𝛼 = 𝛷1 + 𝜃2 − 𝐴 =
45˚, 𝑛𝑔 =
𝐶
𝑉𝑔
, 𝑉𝑔 =
𝐶
𝑛𝑔
=
3∗108
1.414
= 2.12 ∗
108 𝑚
𝑠
.
5- Given: α˳=32̊ , 𝑛𝑝 = 1.732, 𝐴 = 60˚
Request: 𝑛𝑙
Soln: 𝑙𝑛𝑔 =
𝑛𝑔
𝑛𝑙
=
𝛼˳+𝐴
)
2
𝐴
sin( )
2
sin(
= 1.44 =
1.732
𝑛𝑙
𝑆𝑜 𝑛𝑙 = 1.2
6- Given: 𝑛𝑔 = 1.5 + 𝐺𝑟𝑎𝑝ℎ
Request: 𝑇𝑟𝑎𝑐𝑒 𝑡ℎ𝑒 𝑟𝑎𝑦
Soln: sin 𝜑𝑐 =
1
𝑛𝑔
=
1
1.5
𝑆𝑜 𝜑𝑐 = 410 48’
1st face: 𝜑1 = θ1 = 𝑧𝑒𝑟𝑜
2nd face:
𝜑2 = 60 ˂ 𝜑𝑐 𝑡𝑜𝑡𝑎𝑙 𝑖𝑛𝑡𝑒𝑟𝑛𝑎 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑆𝑜 θ2 = 60
3rd face: 𝜑3 = θ3 = 𝑍𝑒𝑟𝑜
1st face: 𝑛1 ∗ 𝑠𝑖𝑛𝜑1 = 𝑛2 ∗ 𝑠𝑖𝑛θ1
𝑠𝑖𝑛45 = 1.5 ∗ 𝑠𝑖𝑛θ1, 𝑆𝑜 θ1 = 28⁰7’
𝐴 = θ1 + 𝜑1 = 45 − 28.13 = 16.87 = 16⁰52’
2nd face: 𝑛1 ∗ 𝑠𝑖𝑛𝜑2 = 𝑛2 ∗ 𝑠𝑖𝑛θ2
1.5 ∗ 𝑠𝑖𝑛16.87 = 1 ∗ 𝑠𝑖𝑛θ2, 𝑆𝑜 θ2 = 25.8 = 25⁰48’
21
1st face: 𝑛1 ∗ 𝑠𝑖𝑛𝜑1 = 𝑛2 ∗ 𝑠𝑖𝑛θ1
1 ∗ 𝑠𝑖𝑛55 = 1.5 ∗ 𝑠𝑖𝑛θ1,𝑆𝑜 θ1 = 33.1 = 33⁰6’
𝐴 = θ1 + 𝜑2 = 60, 𝑆𝑜𝜑2 = 26.9 = 260 54’
2nd face: 𝑛1 ∗ 𝑠𝑖𝑛𝜑2 = 𝑛2 ∗ 𝑠𝑖𝑛θ2
1.5 ∗ 𝑠𝑖𝑛26.9 = 𝑠𝑖𝑛θ2 𝑆𝑜, θ2 = 47.74 = 42⁰44’
1st face: 𝜑1 = θ1 = 450
2nd face: 𝜑2 = θ2 = 45⁰
1st face: 𝜑1 = θ1 = 𝑧𝑒𝑟𝑜
2nd face: 𝜑2 = 600 > 𝜑𝑐, 𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛, 𝛳2 =
600
3rd face: 𝑛1 ∗ 𝑠𝑖𝑛𝜑3 = 𝑛2 ∗ 𝑠𝑖𝑛𝛳3
1.5 ∗ 𝑠𝑖𝑛30 = 𝑠𝑖𝑛θ3 𝑆𝑜 θ3 = 48.6 = 48⁰35’
7- Give: 𝑑 = 4𝑐𝑚 = 0.04𝑚, 𝐴 = 28.3𝑐𝑚2
Request: 𝑛𝑤
Soln:
𝐴 = п ∗ 𝑟 2 = 28.3, 𝑆𝑜 𝑟 = 3𝑐𝑚,
3
𝑡𝑎𝑛𝜑𝑐 = , 𝑆𝑜 𝜑𝑐 = 36.87 = 360 52’
4
𝑛1 ∗ 𝑠𝑖𝑛𝜑𝑐 = 𝑛2 = 1 𝑆𝑜, 𝑛𝑤 = 1.67
8- Given: 𝐴 = 8˚, 𝑛𝑟 = 1.514, 𝑛𝑏 = 1.523
Request: 𝛼˳𝑎𝑣, 𝜔.
Soln:
𝑛𝑦 =
𝑛𝑏+𝑛𝑟
2
= 1.5185, 𝛼˳𝑎𝑣 = 𝐴 ∗ (𝑛𝑦 − 1) = 4.148 = 4˚9ˊ, 𝜔 =
𝑛𝑏−𝑛𝑟
𝑛𝑦−1
= 0.017
9- Given: 𝐴 = 75˚, 𝜃2 = 90, 𝑛𝑔 = √2
Request: 𝛷1
Soln: 𝑛1 ∗ 𝑠𝑖𝑛𝛷2 = 𝑛2 ∗ 𝑠𝑖𝑛𝜃2 = 1, 𝑆𝑜 𝛷2 = 45˚, 𝐴 = 𝛷2 + 𝜃2 = 75, 𝑆𝑜 𝜃2 = 30˚
22
𝑛1 ∗ 𝑠𝑖𝑛𝛷1 = 𝑛2 ∗ 𝑠𝑖𝑛𝜃1 = √2 ∗ 𝑠𝑖𝑛30, 𝑆𝑜 𝛷1 = 45˚
Another Soln: 𝑎𝑡 min 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑛 =
𝑠𝑖𝑛𝜃2
𝑠𝑖𝑛𝛷2
= √2 , 𝛷2 = 45˚
Given: α˳ = 48.5˚, 𝛷1 = 37˚
10-
Request: θ2, 𝐴, 𝑛𝑝
Soln: 𝜃2 = 𝛷1 = 48.5˚, 𝛼˳ = 𝛷1 + 𝜃2 − 𝐴 = 2 ∗ 48.5˚ − 𝐴 = 37, 𝑆𝑜 𝐴 = 60˚,
𝑛𝑝 =
𝑠𝑖𝑛𝛷1 𝑠𝑖𝑛48.5
=
= 1.5
𝑠𝑖𝑛𝜃1 sin (60)
2
Given: 𝑛 = 1.48
11-
Request: 𝑁𝑜 𝑜𝑓 𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑐𝑒𝑠𝑠.
Soln: 𝑛1 ∗ 𝑠𝑖𝑛𝛷1 = 𝑛2 ∗ 𝑠𝑖𝑛𝜃1 = 𝑠𝑖𝑛50 = 1.48 ∗ 𝑠𝑖𝑛𝜃1, 𝑆𝑜 𝜃1 = 31.17˚
1
𝛷𝑐 = sin−1 = 42.51˚, 𝛷2 = 58.83 > 𝛷𝑐 𝑡𝑜𝑡𝑎𝑙 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝑛
𝛷2 = 𝜃2 = 58.83, 𝑋 = 3.1 ∗ 𝑡𝑎𝑛𝜃2 = 5.124𝑐𝑚, 𝑌 = 1.55 ∗ 𝑡𝑎𝑛𝛷1 = 2.56𝑐𝑚
𝑁𝑜 =
12-
42 − 2.56
+1=8
5.124
Given: Table
Sin φ
0
0.342
0.5
0.643
0.766
0.866
0.94
0.985
1
Sin ϴ
x
0.23
0.334
0.43
0.51
0.578
0.627
0.656
y
From the Graph X=0, Y=41.8˚
1.2
1
0.8
0.6
0.4
0.2
0
Φc=41.8˚ @ 90̊, 𝑛 =
0
0.2
0.4
0.6
1
𝑠𝑖𝑛𝛷𝑐
= 1.5
0.8
23