
Refraction of light occurs at the boundary between two different media, such as between glass and air. In refraction, a light ray is bent toward the normal if it passes into a material with a higher index of refraction and away from the normal if the second material has a lower index. A greater difference between the indices of refraction of the two materials causes light to bend more. When light passes from a higher index material (such as water) to a lower index material (such as air), there is a critical angle of incidence. At angles of incidence greater than the critical angle, a light ray is totally reflected back into the incident medium and there is no refracted ray. This is called total internal reflection.

refraction, index of refraction, Snell’s law of refraction, critical angle, total internal reflection


$${n}_{i}\text{\u202f}\mathrm{sin}\text{\u202f}{\theta}_{i}={n}_{r}\text{\u202f}\mathrm{sin}\text{\u202f}{\theta}_{r}$$

 $$\mathrm{sin}\text{\u202f}{\theta}_{c}=\frac{{n}_{r}}{{n}_{i}}$$
 

Review problems and questions 

 Light from air enters a piece of amber (n = 1.55) on a necklace at an angle of 45°. What is the angle of refraction of the light inside the amber?

Answer: The light is at an angle of refraction of 27.1° after entering the amber.
Asked: angle of refraction, θ_{r}, of the light inside the amber Given: angle of incidence, θ_{i} = 45°; index of refraction of amber, n_{r} = 1.55 Relationships: n_{i} sin θ_{i} = n_{r} sin θ_{r}; index of refraction of air, n_{i} = 1.00 Solve: Rearrange to solve the equation for sin θ_{r}:$$\mathrm{sin}{\theta}_{r}=\frac{{n}_{i}\mathrm{sin}{\theta}_{i}}{{n}_{r}}$$Take the inverse sine (sin^{−1}) of both sides and solve for θ_{r}:$${\theta}_{r}={\mathrm{sin}}^{1}\left(\frac{{n}_{i}\mathrm{sin}{\theta}_{i}}{{n}_{r}}\right)={\mathrm{sin}}^{1}\left(\frac{\mathrm{1.00\u202f}\mathrm{sin}\mathrm{\u202f45}\xb0}{1.55}\right)={\mathrm{sin}}^{1}0.456=27.1\xb0$$

 Light passing through glass is completely internally reflected when it hits the glass–air barrier at an angle of incidence of 40.0°. What kind(s) of glass could this possibly be? Use the table of indices of refraction to identify different types of glass.

Answer: The glass could be flint glass or leaded glass.
Asked: types of glass this light could be traveling through Given: refracted medium is air with n_{r} = 1.00; total internal reflection occurs at θ = 40.0°. Relationships: sin θ_{c} = n_{r} ⁄ n_{i}; indices of refraction: window glass n_{w} = 1.52 (window glass), n_{f} = 1.62 (flint glass), n_{l} = 1.7 (leaded glass) Solve: A material can be identified by its index of refraction, n_{i}. Use$$\mathrm{sin}{\theta}_{c}=\frac{{n}_{r}}{{n}_{i}}$$Rearrange and solve for n_{i}, using the given angle as the critical angle:$${n}_{i}=\frac{{n}_{r}}{\mathrm{sin}{\theta}_{c}}=\frac{1.00}{\mathrm{sin}40.0\xb0}=1.56$$The glass has an index of refraction of at least 1.56. Flint and leaded glass each have an index of refraction higher that this, whereas window glass does not.

 Eddie goes into a jewelry store to sell a large diamond cube that has been in the family for generations. The jeweler needs to test the cube to confirm that it is really diamond, but she cannot hit it with a hammer because she might break it. So she tests the block’s ability to refract light. She shines a laser into the block at various angles and measures the refracted angle. Her data are in the table to the right.
 What is the index of refraction for this block?
 What material is this block made out of?
 Using optics, is there a different way to test whether the cube is diamond?


θ_{i}  θ_{r} 
0°  0° 
30.0°  19.6° 
45.0°  28.3° 
60.0°  35.5° 


Answer:  The index of refraction is 1.49.
 According to the table for indexes of refraction, this material is acrylic, not diamond.
 The jeweler could have used the critical angle to identify the material by shining the laser into the block and finding the minimum angle where the light was internally reflected. With its high index of refraction, diamond has a very low critical angle (24.4°) that would be easily recognized.
Solution: Asked: index of refraction n_{r} of the block
Given: angles of incidence (θ_{i}) and refraction (θ_{r}) in the data table Relationships: n_{i} sinθ_{i} = n_{r} sinθ_{r} Solve: $${n}_{i}\mathrm{sin}{\theta}_{i}={n}_{r}\mathrm{sin}{\theta}_{r}$$Rearrange to solve for n_{r}:$${n}_{r}=\frac{{n}_{i}\mathrm{sin}{\theta}_{i}}{\mathrm{sin}{\theta}_{r}}$$Assume the incident medium is air (n_{i} = 1.00). Use any pair of values of (θ_{i}, θ_{r}) from the data table, except the first (0°, 0°) pair. Solve for n_{r}:$${n}_{r}=\frac{{n}_{i}\mathrm{sin}{\theta}_{i}}{\mathrm{sin}{\theta}_{r}}=\frac{\mathrm{1.00\u202f}\mathrm{sin}30.0\xb0}{\mathrm{sin}19.6\xb0}=1.49$$Each pair will produce the same value for n_{r}:$$\begin{array}{l}{n}_{r}=\frac{\mathrm{1.00\u202f}\mathrm{sin}\mathrm{\u202f45.0}\xb0}{\mathrm{sin}28.3\xb0}=1.49\\ {n}_{r}=\frac{\mathrm{1.00\u202f}\mathrm{sin}\mathrm{\u202f60.0}\xb0}{\mathrm{sin}35.5\xb0}=1.49\end{array}$$Answer: The index of refraction of this block is 1.49.  According to the table for indexes of refraction, this material is acrylic, not diamond.
 The jeweler could have used the critical angle to identify the material by shining the laser into the block and finding the minimum angle where the light was internally reflected. With its high index of refraction, diamond has a very low critical angle (24.4°) that would be easily recognized.

 A physicist has three materials (A, B, and C) with indices of refraction 1, 2, and 3, respectively.
 Does a light ray shining from A into B deflect away from or toward the normal?
 Does a light ray shining from C into B deflect away from or toward the normal?
 Does a light ray shining from A into C deflect away from or toward the normal?
 If the angle of incidence is always 30°, does the light ray deflect more in a or in c?

 The ray deflects toward the normal, since B has a higher index of refraction than A.
 The ray deflects away from the normal, since B has a lower index of refraction than C.
 The ray deflects toward the normal, since C has a higher index of refraction than A.
 It deflects more in c. The difference between indices of refraction is higher between A and C than between A and B.

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