Optics—reflection & refraction
Name___________________
Date__________ per.______
1. Reflection: The ray diagram below shows light reflecting from a mirrored surface. Label the
following items: incident ray, reflected ray, interface, normal, angle of incidence, angle of
reflection.
Use a protractor to measure the angle of incidence and reflection.
incident = _______
reflection = ________
Do the rays in this diagram obey the law of reflection? Explain your answer.
Is this an example of regular reflection or diffuse reflection? Explain your answer.
2. Refraction: The ray diagram below shows light traveling from air into a transparent material.
Label the following items: incident ray, reflected ray, transmitted ray, interface, normal, angle
of incidence (1), angle of reflection (1’), angle of refraction (2).
Use your protractor to measure the angle of incidence and the angle of refraction.

1 = ______________
2 = ________________
State Snell’s Law as a formula and use it to determine the index of refraction for the transparent
material.
Calculate the speed of light in the transparent material.
What is the angle between the transmitted ray and the reflected ray?
Suppose the incident ray in this problem was monochromatic with a wavelength of 550 nm.
Find the frequency of the light ray as it travels through the air.
Now state the frequency of the transmitted ray which is traveling through the clear material
(check your notes to remind yourself what happens to the frequency of a wave when it crosses
an interface).
Finally, determine the wavelength of the transmitted portion of the wave.
3. Dispersion: A ray of white light traveling inside of a diamond strikes the diamond-air interface
with an incident angle of 24o. In a diamond the index of refraction is 2.410 for red light, but for
blue light it is 2.450.
Air (n=1.000)
Diamond (nred=2.410
nblue=2.450)
24o
Will the light speed up or slow down as it exits the diamond and enters the air?
Will the ray of light leaving the diamond bend toward or away from the normal?
Use Snell’s Law to calculate the angle of refraction for red light (show your calculation below).
Indicate the ray of red light exiting the diamond in the diagram above (use a protractor to
precisely measure the correct angle).
Use Snell’s Law to calculate the angle of refraction for blue light (show your calculation below.
Indicate the ray of blue light exiting the diamond in the diagram above (use a protractor to
precisely measure the correct angle).
Although the incident ray appears white (owing to the mixing of many different colors of the
spectrum) describe what the light exiting the diamond would look like. What is the name of
this phenomenon? What is an especially beautiful example of this from nature?
4. Critical angle: In the previous problem light rays were bent away from the normal when
traveling from an “optically dense” material (diamond), into a less “optically dense” material
(air). This is another way of saying that the incident ray was traveling slower than the
transmitted ray.
If the incident ray hit the inner surface of the diamond at a larger angle of incidence the
transmitted ray would bend 90o away from the normal so that it was just grazing the surface of
the interface. The angle at which this happens is called the critical angle.
Air (n=1.000)
Diamond(n=2.450)
Transmitted ray
critical
Use Snell’s Law to calculate the critical angle for diamond. (2=90o)
It does not make sense for the angle of refraction (2) to be greater than 90o. Explain why.
Calculate the critical angle for a water-air interface. (nwater=1.33)
5. Total Internal Reflection:
Since no transmitted ray can be bent more than 90o from the normal, any incident angles greater
than critical will experience “total internal reflection”. This simply means that all of the light
reflects back into the material with none of it escaping to the air outside. The inner surface of
the material acts as a perfect mirror for light rays experiencing total internal reflection.
Fiber optics cable, which are used in communications networks and surgical cameras, are an
important technological application of this. Light rays which get sent into the glass fibres of the
cable can bounce around, but can’t escape due to total internal reflection. Thus, the optical
signal can travel great distances (and even around corners) without losing intensity!
air
no escape!!
glass
air
Can total internal reflection occur if a light ray traveling through the air hits a block of
plexiglass? Why or why not?
The critical angle for light coming from a transparent solvent into air is 37o. Determine the
index of refraction of the solvent (remember to set 2=90o to find the critical angle!).
In the space below sketch a ray diagram showing what would happen if a ray of light traveling
through the solvent struck the surface at an angle of incidence equal to 46o. Label your diagram
clearly.
What percentage of the light energy escapes into the air? Explain your answer.