Reflection and refraction

  • Thread starter tigert2004
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  • #1
In Figure 33-48a, a light ray in a liquid (n = 1.61) is incident at angle 1 on a boundary with an underlying material, into which some of the light refracts. There are two choices of underlying material. For each, the angle of refraction 2 versus the incident angle 1 is given in Figure 33-48b. Without calculation, note whether the indices of refraction of material 1 and material 2 are greater or less than the index of the liquid.

What is the index of refraction of material 1?

What is the index of refraction of material 2?

I understand that you use snells law,
and
Material #1:
Incident Angle = 67.5 deg
Refraction Angle = 56.25 deg
→ n1 = nwater*Sin(Incid Ang)/Sin(Refr Ang)
= (1.61)*Sin(67.5 deg)/Sin(56.25 deg)
= 1.79

which isn't correct
 

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  • #2
The picture is still pending approval... but my gut instinct is that you may have your angles mixed up. The angle stays with the index of the material that ray is in:

[tex]n_{air}\sin{\theta_{air}}=n_{water}\sin{\theta_{water}}[/tex]
 
  • #3
OlderDan
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tigert2004 said:
In Figure 33-48a, a light ray in a liquid (n = 1.61) is incident at angle 1 on a boundary with an underlying material, into which some of the light refracts. There are two choices of underlying material. For each, the angle of refraction 2 versus the incident angle 1 is given in Figure 33-48b. Without calculation, note whether the indices of refraction of material 1 and material 2 are greater or less than the index of the liquid.

What is the index of refraction of material 1?

What is the index of refraction of material 2?

I understand that you use snells law,
and
Material #1:
Incident Angle = 67.5 deg
Refraction Angle = 56.25 deg
→ n1 = nwater*Sin(Incid Ang)/Sin(Refr Ang)
= (1.61)*Sin(67.5 deg)/Sin(56.25 deg)
= 1.79

which isn't correct
Try using a different point on the graph where the curve intersects the grid points on the right edge. It will make some difference in your answer.
 

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