<< Chapter < Page Chapter >> Page >
Refractive indices of some materials. n air is calculated at STP.
Medium Refractive Index
Vacuum 1
Helium 1,000036
Air* 1,0002926
Carbon dioxide 1,00045
Water: Ice 1,31
Water: Liquid ( 20 C) 1,333
Acetone 1,36
Ethyl Alcohol (Ethanol) 1,36
Sugar solution (30%) 1,38
Fused quartz 1,46
Glycerine 1,4729
Sugar solution (80%) 1,49
Rock salt 1,516
Crown Glass 1,52
Sodium chloride 1,54
Polystyrene 1,55 to 1,59
Bromine 1,661
Sapphire 1,77
Glass (typical) 1,5 to 1,9
Cubic zirconia 2,15 to 2,18
Diamond 2,419
Silicon 4,01

Snell's law

Now that we know that the degree of bending, or the angle of refraction, is dependent on the refractive index of a medium, how do we calculate the angle of refraction?

The angles of incidence and refraction when light travels from one medium to another can be calculated using Snell's Law.

Snell's Law
n 1 sin θ 1 = n 2 sin θ 2

where

n 1 = Refractive index of material 1
n 2 = Refractive index of material 2
θ 1 = Angle of incidence
θ 2 = Angle of refraction

Remember that angles of incidence and refraction are measured from the normal, which is an imaginary line perpendicular to the surface.

Suppose we have two media with refractive indices n 1 and n 2 . A light ray is incident on the surface between these materials with an angle of incidence θ 1 . The refracted ray that passes through the second medium will have an angle of refraction θ 2 .

A light ray with an angle of incidence of 35 passes from water to air. Find the angle of refraction using Snell's Law and [link] . Discuss the meaning of your answer.

  1. From [link] , the refractive index is 1,333 for water and about 1 for air. We know the angle of incidence, so we are ready to use Snell's Law.

  2. According to Snell's Law:

    n 1 sin θ 1 = n 2 sin θ 2 1 , 33 sin 35 = 1 sin θ 2 sin θ 2 = 0 , 763 θ 2 = 49 , 7 or 130 , 3

    Since 130 , 3 is larger than 90 , the solution is:

    θ 2 = 49 , 7
  3. The light ray passes from a medium of high refractive index to one of low refractive index. Therefore, the light ray is bent away from the normal.

Got questions? Get instant answers now!

A light ray passes from water to diamond with an angle of incidence of 75 . Calculate the angle of refraction. Discuss the meaning of your answer.

  1. From [link] , the refractive index is 1,333 for water and 2,42 for diamond. We know the angle of incidence, so we are ready to use Snell's Law.

  2. According to Snell's Law:

    n 1 sin θ 1 = n 2 sin θ 2 1 , 33 sin 75 = 2 , 42 sin θ 2 sin θ 2 = 0 , 531 θ 2 = 32 , 1 .
  3. The light ray passes from a medium of low refractive index to one of high refractive index. Therefore, the light ray is bent towards the normal.

Got questions? Get instant answers now!

If

n 2 > n 1

then from Snell's Law,

sin θ 1 > sin θ 2 .

For angles smaller than 90 , sin θ increases as θ increases. Therefore,

θ 1 > θ 2 .

This means that the angle of incidence is greater than the angle of refraction and the light ray is bent toward the normal.

Similarly, if

n 2 < n 1

then from Snell's Law,

sin θ 1 < sin θ 2 .

For angles smaller than 90 , sin θ increases as θ increases. Therefore,

θ 1 < θ 2 .

This means that the angle of incidence is less than the angle of refraction and the light ray is away toward the normal.

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Siyavula textbooks: grade 10 physical science. OpenStax CNX. Aug 29, 2011 Download for free at http://cnx.org/content/col11245/1.3
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Siyavula textbooks: grade 10 physical science' conversation and receive update notifications?

Ask