Lecture 18: Interactions of Light with Nonconductors

OCW Scholar

« Previous | Next »

Lecture Topics

A cartoon image of an astronaut walking in space.
  • Reflection and Refraction at Interfaces of Dielectrics
  • Total Internal Reflection
  • Fresnel's Equations
  • Amplitude and Intensity
  • Polarization Effects in Reflection

Learning Objectives

By the end of this lecture, you should:

  • derive reflection and refraction conditions at a dielectric interface.
  • find the critical angle and describe total internal reflection.
  • understand how boundary conditions are applied at dielectric interfaces.
  • use Fresnel's equations to solve reflection/transmission problems.
  • find the intensity of light in reflection/transmission problems.
  • find the Brewster angle.
  • discuss polarization effects in reflection and refraction involving dielectrics.

Lecture Activities

Check Yourself

  • In an application where it is needed to bend light through 90º, a telescope manufacturer is considering whether to use a 45º glass prism with total internal reflection, or reflection from an aluminum layer deposited on glass (held at a 45º angle) which gives 92% reflection. Discuss, with rough numbers, the merits of each approach based on what you have learned.

View/hide answer

Quiz18.jpgIn the prism shown, about 4% of the light would be lost to reflection at the entry point of air to glass, and a similar amount at the exit surface. The total internal reflection will reflect 100% of the light in the plane perpendicular to the page, and slightly less in the plane parallel to the page. So overall there will be a bit less light than the 92% the aluminum would have given, and it will be slightly polarized. The final decision might rest on other factors like the glass surfaces of a prism being more resistant to abrasion than an aluminum layer is, or what manufacturing facilities are available.


  • A Brewster plate is often used as a window on a high-powered gas laser. This is very similar to a single glass plate selected from the stcked plated used in the lecture. Consider initially unpolarized light to go from air into n=1.5 glass at the Brewster angle, move about 1 mm in the glass, and then go through a glass to plasma (index very close to n=1 like air). Do not consider multiple reflections, and refer to the lecture for the result of Fresnel’s equations. What will be the intensity and polarization state of the transmitted light after one pass through such a plate?

View/hide answer

At the Brewster angle, 100% of the light polarized parallel to the glass will enter. It will enter the other wall of the glass at the glass-to-air Brewster angle, and again the parallel polarized light will pass 100%. By Snell’s Law, the same angles are involved in exiting, except for a sign, so the transmission of 85% (100% - 15%) of the perpendicularly polarized light applies each time for .852 = .7225 of it getting through. The light has 100% of the original parallel component and 72% of the original perpendicular component. In a gas laser, multiple passes through such plates rapidly results in 100% parallel polarization.


  • Sunlight coming onto a smooth water surface when the Sun is 2º above the horizon is reflected from the water. What is the intensity of the reflected beam compared to the sunbeam itself? What relative intensity will get through if it is viewed through a perfect polarizer oriented vertically?

View/hide answer

First we must use Snell's Law to find the refraction angle \({\theta _2}\). Since \({n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}\), we have with the appropriate indices \[{\theta _2} = {\sin ^{ - 1}}\left( {\frac{{{n_1}\sin {\theta _1}}}{{{n_2}}}} \right) = {\sin ^{ - 1}}\left( {\frac{{1\sin 88^\circ }}{{1.33}}} \right) = 48.7^\circ \] basically the same result as for 90º incidence in the lecture. We can solve Fresnel’s equations for the amplitude ratios:\[{r_\parallel } = - \frac{{\tan ({\theta _1} - {\theta _2})}}{{\tan ({\theta _1} + {\theta _2})}} = - \frac{{\tan (88 - 48.7)}}{{\tan (88 + 48.7)}} = - \frac{{\tan (39.3)}}{{\tan (136.7)}} = \frac{{.81849}}{{.94235}} = .8686\]and \[{r_ \bot } = - \frac{{\sin ({\theta _1} - {\theta _2})}}{{\sin ({\theta _1} + {\theta _2})}} = - \frac{{\sin (39.3)}}{{\sin (136.7)}} = - \frac{{.63338}}{{.68582}} = .9235\] To get intensities we need to square these to find that the parallel light is 75% and the perpendicular light 85% the intensity of the incoming beam. The vertically oriented polarizer would cut the parallel component, or less than half of the incoming light. Polarizing sunglasses would be relatively ineffective at cutting this sort of glare. (With the Sun higher, near the Brewster angle, they would be very effective.)


« Previous | Next »