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Comment: Triangulation is a process that can be applied to solve problems in a number of areas of engineering including surveying, construction management, radar, sonar, lidar, etc.


Refraction is a physical phenomenon that occurs when light passes from one transparent medium (such as air) through another (for example, glass.) It is known that light travels at different speeds through different transparent media. The index of refraction of a medium is a measure of how much the speed of light is reduced as it passes through the medium. In the case of glass, the index of refraction is approximately 1.5. This means that light travels as a speed of 1 1 . 5 = 2 3 size 12{ { {1} over {1 "." 5} } = { {2} over {3} } } {} times the speed of light in a vacuum.

Two common properties of transparent materials can be attributed to the index of refraction. One is that light rays change direction as they pass from one medium through another. Secondly, light is partially reflected when it passes from one medium to another medium with a different index of refraction. We will focus on the first of these properties in this reading.

In optics, which is a field of physics, you will learn about Snell's law, which is also known as Descartes' law after the scientist, Rene Descartes . Snell’s law takes the form of an equation that states the relationship between the angle of incidence and the angle of refraction for light passing from one medium to another. Stated mathematically, Snell’s law is

sin ( θ incidence ) sin ( θ refraction ) = c incidence c refraction size 12{ { {"sin" \( θ rSub { size 8{ ital "incidence"} } \) } over {"sin" \( θ rSub { size 8{ ital "refraction"} } \) } } = { {c rSub { size 8{ ital "incidence"} } } over {c rSub { size 8{ ital "refraction"} } } } } {}

It follows that

sin ( θ incidence ) sin ( θ refraction ) = I 2 I 1 size 12{ { {"sin" \( θ rSub { size 8{ ital "incidence"} } \) } over {"sin" \( θ rSub { size 8{ ital "refraction"} } \) } } = { {I rSub { size 8{2} } } over {I rSub { size 8{1} } } } } {}

where I 1 and I 2 size 12{I rSub { size 8{1} } ` ital "and"`I rSub { size 8{2} } } {} are the Index of Refraction of medium 1 and medium 2 respectively.

Consider a situation where light rays pass are shined from air through a tank of water. This situation is illustrated below.

Depiction of light refraction.

The Index of refraction for air is 1.0003 and that of water is 1.3000. Let us assume that the angle that light enters the water is 21 0 40’, what is the angle of refraction, w ?

From Snell’s law, we know

I W I A = sin ( a ) sin ( w ) size 12{ { {I rSub { size 8{W} } } over {I rSub { size 8{A} } } } = { {"sin" \( a \) } over {"sin" \( w \) } } } {}
sin ( w ) = I A I W sin ( a ) size 12{"sin" \( w \) = { {I rSub { size 8{A} } } over {I rSub { size 8{W} } } } `"sin" \( a \) } {}

sin ( w ) = I A I W sin ( a ) size 12{"sin" \( w \) = { {I rSub { size 8{A} } } over {I rSub { size 8{W} } } } `"sin" \( a \) } {}

Substituting in the numerical values for I A , I W and a yield

sin ( w ) = 0 . 2841 size 12{"sin" \( w \) =0 "." "2841"} {}

We now make use of the inverse sine function

w = sin 1 ( 0 . 2841 ) size 12{w="sin" rSup { size 8{ - 1} } \( 0 "." "2841" \) } {}

This leads to the result

w = 16 0 30 ' size 12{w="16" rSup { size 8{0} } `"30" rSup { size 8{'} } } {}

We conclude that the refracted ray will travel through the water at an angle of refraction of 16 0 30’ .


  1. A 50 ft ladder leans against the top of a building which is 30 ft tall. Determine the angle the ladder makes with the horizontal. Also determine the distance from the base of the ladder to the building.
  2. A straight trail leads from the Alpine Hotel at elevation 8,000 feet to a scenic overlook at elevation 11,100 feet. The length of the trail is 14,100 feet. What is the inclination angle β in degrees? What is the value of β in radians?
  3. A ray of light moves from a media whose index of refraction is 1.200 to another whose index of refraction is 1.450. The angle of incidence of the ray as it intersects the interface of the two media is 15 0 . Sketch the geometry of the situation and determine the value of the angle of refraction.
  4. One-link planar robots can be used to place pick up and place parts on work table. A one-link planar robot consists of an arm that is attached to a work table at one end. The other end is left free to rotate about the work space. If l = 5 cm, sketch the position of the robot and determine the ( x, y ) coordinates of point p ( x , y ) for the following values for θ: (50˚, 2π/3 rad, -20˚, and -5 π/4 rad).

One-link planar robot.

Questions & Answers

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s. Reply
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of graphene you mean?
or in general
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Graphene has a hexagonal structure
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
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Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
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