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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.

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.

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.

Both these situations can be seen in [link] .

Refraction of two light rays. (a) A ray travels from a medium of low refractive index to one of high refractive index. The ray is bent towards the normal. (b) A ray travels from a medium with a high refractive index to one with a low refractive index. The ray is bent away from the normal.

What happens to a ray that lies along the normal line? In this case, the angle of incidence is 0 and

sin θ 2 = n 1 n 2 sin θ 1 = 0 θ 2 = 0 .

This shows that if the light ray is incident at 0 , then the angle of refraction is also 0 . The ray passes through the surface unchanged, i.e. no refraction occurs.

Investigation : snell's law 1

The angles of incidence and refraction were measured in five unknown media and recorded in the table below. Use your knowledge about Snell's Law to identify each of the unknown media A - E. Use [link] to help you.

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Source:  OpenStax, Maths test. OpenStax CNX. Feb 09, 2011 Download for free at http://cnx.org/content/col11236/1.2
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