# Geometric Optics

## Contents

#### Problems on Refraction

Problem : As light moves from air ( n 1.00 ) to amber it deviates 18 o from its 45 o angle of incidence. Which way does it bend? What is the speed of light in amber?

Light entering a denser medium refracts towards the normal. Thus the angle of refraction is θ t = 45 - 18 = 27 o . Using Snell's Law we have n t = 1.56 . The speed in amber is given by v = c/n = 3.0×108/1.56 1.92×108 m/s or 0.64c .

Problem : A transparent fiber of index of refraction 1.6 is surrounded (cladded) by a less dense plastic of index 1.5. At what angle must a light ray in the fiber approach the interface so as to remain within the fiber?

This problem involves total internal reflection. The critical angle for staying within the fiber is given by: sinθ c = = 1.5/1.6 = 0.938 . Thus θ c = 69.6 o . The ray must approach the interface between the media at an angle of 69.6 o or greater to the normal.

Problem : A light ray in air approaches a water surface ( n 1.33 ) such that its electric vector is parallel to the plane of incidence. If θ i = 53.06 o , what is the relative amplitude of the reflected beam? What about if the electric field is perpendicular to the plane of incidence?

We can apply the Fresnel Equations. In the first case we want the expression for r || . From Snell's Law we can deduce that sinθ t = (n i/n t)sinθ i which implies θ t = 36.94 o . Then:

 r || = 0

In the latter (perpendicular) case we have

 r âä¥ = = - 0.278

In the former case, no light is reflected -- this is called Brewster's angle as we shall see in the section on polarization. For the perpendicular field the amplitude of the reflected wave is 0.278 as large as the incident wave. That is the reflected ray is about (0.278)2 0.08 , or about 8% as bright as the incident ray (irradiance is proportional to the square of the amplitude).

Problem : By what angle do blue light ( λ b = 460 nm) and red light ( λ r = 680 nm) disperse upon entering (from vacuum) a medium with N = 7×1038 , ε = 1.94 , and σ 0 = 5.4×1015 Hz at an incident angle of 20 o (the electron charge is 1.6×10-19 Coulombs and its mass is 9.11×10-31 kilograms)?

First we must calculate the index of refraction for both light frequencies. The angular frequency of the blue light is σ b = 4.10×1015 Hz and for the red light σ r = 2.77×1015 . Thus we have:

 n r 2 = 1 + = 1 + = 1 + 0.472

Thus n r = 1.213 . Similarly for the blue:

 n b 2 = 1 + = 1 + = 1 + 0.821

Thus n b = 1.349 . We can then calculate the angles of refraction of the two beams as they enter the medium from Snell's Law. For the red: 1.213 sinθ r = sinθ i . This gives θ r = sin-1(sin(20 o )/1.213) = 16.38 o . For the blue: 1.349 sinθ b = sinθ i . Giving: θ b = 14.69 o . The difference between these two angles is 1.69 o , which is the amount by which the different colored rays disperse.

Problem : What is the rate of change of the speed of light with angular frequency in a dielectric medium?

We want . But v = c/n . So since c is a constant, we can say = c . Now:

 = âá’1n =

Now we must take the derivative of this with respect to σ treating everything else as a constant. This gives:

 =

The required derivative is just c times this:

 =