Characteristics of Waves

Physics 11 Name:______Blk: _____ Chapter 16, 17 Lesson 3 The Critical Angle

Index of Refraction

 To understand the behavior of light in different properties, we refer to the index of refraction n.  It is the ratio of the speed of light in a vacuum, c, to the speed of light in a given material, v.  Mathematically, c n  v  The higher the index of refraction, the more light is slowed down when it travels from a vacuum in to a substance.

Examples 1. The speed of light in a liquid is 2.25  108 m/s. What is the refractive index of the liquid?

2. Calculate the speed of light in Lucite (Plexiglas), if nlucite = 1.51

Indices of Refraction

Substance Index of Refraction ( n ) Vacuum 1.0000 Air (0C, 101 kPa) 1.0003 Water 1.33 Ethyl alcohol 1.36 Quartz (fused) 1.46 Glycerin 1.47 Lucite or Plexiglas 1.51 Glass (crown) 1.52 Sodium chloride 1.53 Glass (crystal) 1.54 Ruby 1.54 Glass (flint) 1.65 Zircon 1.92 Diamond 2.42 Note: For yellow light, wavelength = 589 nm

Using Snell's Law to Calculate the Critical Angle if a ray of light is coming from air into water, Snell's Law might be applied, and = nw, where nw is the index of refraction of water.

Reflectance of light between two medium for various angles.

When light moves through media, some light is reflected, some is refracted. As the angle of incidence increases, the angle of refraction will increase...

There comes a point where the angle of refraction reaches ______! Beyond this, refraction stops! Since the light cannot be refracted out of the medium, it is reflected internally!

The angle of incidence that produces 90 degree angle of refraction is unique to different medium. It is called the ______. Angles less than the critical will result in refraction... greater than: ______.

Since the critical angle comes when the angle of refraction is 90 degrees, there is a simple way of calculating it!... Since sin  (refraction) = sin (90) = ______, then Snell’s Law reduces easily: sin  n i = r sin r ni sin  sin  n c = c = r sin(90) 1 ni So from a medium into air... sin  1 c = 1 ni 1 sin c = ni Example 1: Determine the critical angle of a) Water (n= 1.33)

b) Glass (n=1.52)

c) diamond (n=2.42)

Example 2: What is the critical angle of light traveling from a diamond to a ruby? Total Internal Reflection & Refraction Exercise

Here we have a ray of light entering two cut stones: a diamond (n = 2.42) and a cubic zirconium (n = 1.92). Can you tell the difference?

Use your knowledge of the law of reflection, the critical angle, and total internal reflection to determine the path of the light inside the stone, and where it will eventually exit. You must use a protractor and ruler to do this properly

Be sure to label each angle measurement and show the calculations for the critical angle. DIAMOND (n = 2.42)

135º 135º

90º 90º

Physics 11 Name: Lab – Refraction Date: Materials:  Rectangular “glass” block  Ray box  Protractor  Brain cells

Purpose: To determine the relationship between the angle of incidence and the angle of refraction.

Diagram: Normal line Angle of Incidence

Angle of Refraction

Procedure: Direct a single ray of light at the side of the glass block (called the incident ray) as shown above. Place the ray box to create an angle of incidence of 5˚ and record it. Measure and record the angle of refraction. Note that the bent beam of light is called the refracted ray. Increase the angle of incidence by 5˚ increments and repeat your measurements. Do this until there are no longer refracted rays coming out of the other side of the glass block. Graph the results by plotting sin i versus sin r (put sin r on the x axis).

Data: Angle of Angle of sin i sin r sin i Incidence, i Refraction, r sin r

Conclusion: