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25.3 The law of refraction  (Page 4/7)

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The law of refraction

n 1 sin θ 1 = n 2 sin θ 2 size 12{n rSub { size 8{1} } "sin"θ rSub { size 8{1} } =n rSub { size 8{2} } "sin"θ rSub { size 8{2} } } {}

Take-home experiment: a broken pencil

A classic observation of refraction occurs when a pencil is placed in a glass half filled with water. Do this and observe the shape of the pencil when you look at the pencil sideways, that is, through air, glass, water. Explain your observations. Draw ray diagrams for the situation.

Determine the index of refraction from refraction data

Find the index of refraction for medium 2 in [link] (a), assuming medium 1 is air and given the incident angle is 30 . 0º size 12{"30" "." 0°} {} and the angle of refraction is 22 . 0º size 12{"22" "." 0°} {} .

Strategy

The index of refraction for air is taken to be 1 in most cases (and up to four significant figures, it is 1.000). Thus n 1 = 1 . 00 size 12{n rSub { size 8{1} } =1 "." "00"} {} here. From the given information, θ 1 = 30 . 0º size 12{q rSub { size 8{1} } ="30" "." 0°} {} and θ 2 = 22 . 0º size 12{q rSub { size 8{2} } ="22" "." 0°} {} . With this information, the only unknown in Snell’s law is n 2 size 12{n rSub { size 8{2} } } {} , so that it can be used to find this unknown.

Solution

Snell’s law is

n 1 sin θ 1 = n 2 sin θ 2 . size 12{n rSub { size 8{1} } "sin"θ rSub { size 8{1} } =n rSub { size 8{2} } "sin"θ rSub { size 8{2} } } {}

Rearranging to isolate n 2 size 12{n rSub { size 8{2} } } {} gives

n 2 = n 1 sin θ 1 sin θ 2 . size 12{n rSub { size 8{2} } =n rSub { size 8{1} } { {"sin"θ rSub { size 8{1} } } over {"sin"θ rSub { size 8{2} } } } } {}

Entering known values,

n 2 = 1 . 00 sin 30 . 0º sin 22 . 0º = 0 . 500 0 . 375 = 1.33. alignl { stack { size 12{n rSub { size 8{2} } =1 "." "00" { {"sin""30" "." 0°} over {"sin""22" "." 0°} } = { {0 "." "500"} over {0 "." "375"} } } {} #=1 "." "33" "." {} } } {}

Discussion

This is the index of refraction for water, and Snell could have determined it by measuring the angles and performing this calculation. He would then have found 1.33 to be the appropriate index of refraction for water in all other situations, such as when a ray passes from water to glass. Today we can verify that the index of refraction is related to the speed of light in a medium by measuring that speed directly.

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A larger change in direction

Suppose that in a situation like that in [link] , light goes from air to diamond and that the incident angle is 30 . 0º size 12{"30" "." 0°} {} . Calculate the angle of refraction θ 2 size 12{q rSub { size 8{2} } } {} in the diamond.

Strategy

Again the index of refraction for air is taken to be n 1 = 1 . 00 size 12{n rSub { size 8{1} } =1 "." "00"} {} , and we are given θ 1 = 30 . 0º size 12{q rSub { size 8{1} } ="30" "." 0°} {} . We can look up the index of refraction for diamond in [link] , finding n 2 = 2 . 419 size 12{n rSub { size 8{2} } =2 "." "419"} {} . The only unknown in Snell’s law is θ 2 size 12{q rSub { size 8{2} } } {} , which we wish to determine.

Solution

Solving Snell’s law for sin θ 2 size 12{q rSub { size 8{2} } } {} yields

sin θ 2 = n 1 n 2 sin θ 1 . size 12{"sin"θ rSub { size 8{2} } = { {n rSub { size 8{1} } } over {n rSub { size 8{2} } } } "sin"θ rSub { size 8{1} } } {}

Entering known values,

sin θ 2 = 1 . 00 2 . 419 sin 30 . 0º = ( 0 . 413 ) ( 0 . 500 ) = 0 . 207 . size 12{"sin"q rSub { size 8{2} } = { {1 "." "00"} over {2 "." "419"} } "sin""30" "." 0"°=" left (0 "." "413" right ) left (0 "." "500" right )=0 "." "207"} {}

The angle is thus

θ 2 = sin − 1 0 . 207 = 11 . 9º . size 12{θ rSub { size 8{2} } ="sin" rSup { size 8{ - 1} } 0 "." "207"="11" "." 9°} {}

Discussion

For the same 30º angle of incidence, the angle of refraction in diamond is significantly smaller than in water ( 11.9º rather than 22º —see the preceding example). This means there is a larger change in direction in diamond. The cause of a large change in direction is a large change in the index of refraction (or speed). In general, the larger the change in speed, the greater the effect on the direction of the ray.

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Section summary

  • The changing of a light ray’s direction when it passes through variations in matter is called refraction.
  • The speed of light in vacuum c = 2 . 9972458 × 10 8 m/s ≈ 3 . 00 × 10 8 m/s . size 12{c=2 "." "9972458" times "10" rSup { size 8{8} } " m/s" approx 3 "." "00" times "10" rSup { size 8{8} } " m/s"} {}
  • Index of refraction n = c v size 12{n= { {c} over {v} } } {} , where v size 12{v} {} is the speed of light in the material, c size 12{c} {} is the speed of light in vacuum, and n size 12{n} {} is the index of refraction.
  • Snell’s law, the law of refraction, is stated in equation form as n 1 sin θ 1 = n 2 sin θ 2 size 12{n rSub { size 8{1} } "sin"θ rSub { size 8{1} } =n rSub { size 8{2} } "sin"θ rSub { size 8{2} } } {} .

Conceptual questions

Diffusion by reflection from a rough surface is described in this chapter. Light can also be diffused by refraction. Describe how this occurs in a specific situation, such as light interacting with crushed ice.

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Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Kate Reply
what is the change in momentum of a body?
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temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
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specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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