ALSO, attempt, but do not hand in, the mock final exam. (I will go over the answers to this exam on Monday March 22.)

Question 21.8

Can a material have a refractive index less than unity? Explain.

Answer 21.8

No, it can't. The refractive index is the ratio of the speed of light in vacuum to the speed of light in the medium, and nothing travels faster than the speed of light in vacuum. (With the possible exception of the hypothetical particles known as tachyons.)

Note added March 31, 2004: after looking into this a bit further, I find that a material CAN have a refractive index less than unity; in fact, most materials have this property when exposed to X-rays. And this does mean that the effective speed of the rays in the material is greater than the speed of light in vacuum, although it remains impossible to transmit information faster than light. There's a good discussion of this in the Feynman Lectures in Physics, Volume I, Chapter 31.

(My guess is that Callister was expecting the answer `No', though.)

Question 21.13

We want the reflectivity of light at normal incidence to the surface of a tranparent medium to be less than 5%. Which of the following materials listed in Table 21.1 are likely candidates: soda-lime glass, Pyrex glass, periclase, spinel, polystyrene and polypropylene?

Answer 21.13

We assume that the transparent medium is in air. The refractive index of air may be taken as 1. The reflectivity of a beam normal to the surface of a medium with refractive index n in air is given by Equation 21.13:

R = ((n - 1)/(n + 1))²

So we solve for R = 0.05:

Taking square roots of each side, we get:

0.2236 =(n - 1)/(n + 1) [ignoring the negative roots].

From which we obtain that:

n = (1 + 0.2236)/(1 - 0.2236) = 1.576

So this is the index that would give a reflectivity of exactly 5%. If we increase the index, the reflectivity will increase, so we are looking for materials with an index less than 1.576; soda-lime glass, pyrex and polypropylene would all be candidates.

Question 21.20

Derive Equation 21.19, starting from other expressions given in the Chapter.

Answer 21.20

Equation 21.19 gives the intensity of the beam transmitted through a parallel-sided slab of a material of thickness l, adsorbtion coefficient beta and reflectivity R:

I_t = I₀(1-R)²e^{-beta * l}

To obtain this equation, we first consider reflection at the front surface of the slab. A fraction I₀R of the incident beam is reflected at this surface, so the intensity of the beam that enters the medium is I₀(1-R). This beam is then attenuated as it passes through the medium according to Equation 21.18, so when it gets to the back surface of the slab, it has intensity I₀(1-R)e^{-beta * l}. At the back surface, a fraction I₀(1-R)e^{-beta * l}R is reflected back into the slab, so the intensity of the beam emerging from the slab is I₀(1-R)²e^{-beta * l}, as required.

Question 21.27

In your own words, describe briefly the phenomenon of photoconductivity. Would the semiconductor zinc selenide, which has a band gap of 2.58 eV, be photoconductive when exposed to visible light? Why, or why not?

Answer 21.27

Photoconductivity is found in insulators or semi-conductors having a band gap between the valence band and the conduction band with an energy corresponding to a photon of visible light. When a photon of the right wavelength strikes the material, it moves an electron up from the valence band to the conduction band. Both the electron and the hole it leaves behind in the valence band then act as charge carriers, making the material conductive.

If the band gap in zinc selenide is 2.58 eV, this corresponds to a photon of frequency nu, where

2.58 = 6.63 * 10^-34 * nu / (1.602 * 10^-19)

From which nu = 0.6234 * 10¹⁵ Hz

The wavelength corresponding to this frequency is lambda = c/nu meters

From which lambda = 3 * 10⁸/(0.6234 * 10¹⁵) = 4.81 * 10^-7 m

= 0.481 microns. This corresponds to blue light, so zinc selenide will be photoconductive under blue light (though not under longer wavelengths such as red.)