A refl = R A inc, A trans = T A inc.

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1 Reading: Wave Optics 1, 2 Key concepts: Superposition; phase difference; amplitude and intensity; thin film interference; Fraunhofer diffraction; gratings; resolving power. 1.! Questions about interference of waves. Why do we not see interference patterns in the light from the two headlights of a distant car? You are standing outdoors near a corner of a building. Two people around the corner are talking. You can hear them but not see them. How do you hear them? [Reflection from other objects is negligible.] You know the two people are talking, but you are unable to understand their conversation. Why? [They are speaking your language.] 2.! Questions about thin films. When a film of water on glass evaporates the reflected light becomes very bright just before the film is gone. Why?! When a soap film in air gets thin just before breaking the reflected light is very dim. Why? A circular film of oil on water is thinnest near the edge. In the reflected light, would one see blue or red colors in that region of the film? 3.! In Assignment 8 you examined the intensities of the reflected and transmitted rays shown. To analyze the interference of the waves you must work with the amplitudes. Since I A 2 The relations between amplitude magnitudes are: A refl = R A inc, A trans = T A inc. A 0 a b c Show that if R << 1 then we have A b A a >> A c. [This is why we considered only waves a and b in analyzing interference in the reflected light from a thin film.] a b c Show that in this case A a A 0 >> A b >> A c. [This is why interference effects in the transmitted light are negligible for ordinary films.] Show that if R 1 we have A a A 0 >> A b >> A c, so usually nearly all the light is reflected and interference is (usually) negligible.

2 4.! The exception to the conclusion in 3(c) occurs when all the transmitted waves interfere constructively. Write these amplitudes in terms of A 0 and R. Assuming all these waves are in phase, find the total transmitted amplitude. Use the fact that if x < 1 then 1 + x + x 2 + x 3 + = 1 1 x. What does this say about the transmitted light? What about the reflected light?! 5.! We examine the case R 1 in more detail. The transmitted waves have (complex) amplitudes E a = A 0 T, E b = A 0 T Re iδ, E c = A 0 T R 2 e 2iδ,! and so on. Here δ is the phase difference caused by the wave traveling back and forth between the surfaces; thus δ = 2nkt, where n is the film s refractive index and t is its thickness. Show that I trans I 0 = (1 R) R 2 2Rcosδ. [Remember eiδ + e iδ = 2cosδ.] Let R = Use a graphing calculator to plot this formula as a function of δ for π δ 3π. Note the extremely narrow peak at δ = 2π. In (b) take n = 1.5 and t = 200 nm. For what visible wavelength will there be nearly total transmission? If you look at an incident beam of white light transmitted through this system, what will you see? 6.! If light is interpreted as a stream of particle-like photons, the spread δθ in the directions they move is interpreted as a spread δ p in the component p of their momentum (p) perpendicular to the original direction of the beam. That is, before they went through the circular opening they were all moving in the same direction with p = 0, but afterwards their directions vary within the range δθ. Approximately we have δ p = p δθ. The momentum of a photon is related to its wavelength by the quantum formula p = h/λ, where h is Planck s constant. Show that D δ p 1.22h. [This is an example of the Uncertainty Principle.]

3 7.! A wedge-shaped air film is produced by slipping a object of thickness t between two flat glass plates as shown. Light with λ = 500 nm is normally incident from above. In the reflected light there are 400 bright interference bands across the length of the plate, with a dark band at the end where the object is inserted. The plates are 10 cm long and the glass has refractive index 1.4. light t What is t? Ans: 0.1 mm. If the region between the plates were filled with oil with n = 1.25, how many bright bands would be seen? Ans: 500. Suppose the plates are 20 cm long, but the same object is between them as shown. How would this affect the pattern observed (without the oil)? 8.! A lens is coated so as to reduce reflection in the infrared and ultraviolet regions. The coating causes destructive interference in the reflected light for wavelengths 1080 nm and 360 nm but no wavelength in between. If the coating has refractive index 1.4 and the lens has refractive index 1.6, what is the minimum thickness of the coating? Ans: 193 nm. For what visible wavelength is there constructive interference in the reflected light? Ans: 540 nm. 9.! Consider the coated lens in #10, with light normally incident from air. Calculate the reflectivity at the air-coating interface and at the coating-lens interface. Call these R 1 and R 2, respectively. Consider interference between the waves reflected from the coating and from the lens. If it is constructive, what fraction of the incident intensity is reflected? [What are the amplitudes of the two waves in terms of R 1 and R 2? [Take T 1 at the air-coating interface to be 1.] What fraction is reflected if the interference is destructive? [It is not zero.]

4 10.! Questions about resolving power. The resolution of an optical microscope can be increased by immersing the objective lens and the sample in a transparent oil. Why does this work? Radio telescopes make images of distant objects using the waves emitted in the radio part of the spectrum. If a radio telescope receiving waves with λ = 3 cm is to have the same resolution the 5 m (diameter) Hale telescope has with 500 nm light, how large must the diameter of the antenna be? The electrons in an electron microscope are accelerated through a potential difference sufficient to give them a momentum p = mc. [This does not mean they are traveling at speed c; the formula p = mv is valid only for speeds much smaller than c.] Find their debroglie wavelength from the formula λ = h/p, where p is the momentum and h is Planck s constant. Ans: nm. 11.! Questions about gratings. A grating s ability to resolve two spectral lines with nearly the same wavelength is improved if the intensity peaks are narrower. Which factor in the formula for resolving power R = mn expresses this fact? This ability is also improved if the peaks are farther apart in angle. Which factor express this fact? Show that no matter how many lines a grating has, the visible spectrum in 3rd order always overlaps that in 2nd order. 12.! A grating 2 cm wide is used to determine wavelengths of visible light. If two wavelengths with λ av = 500 nm and Δλ = nm are resolved in 2nd order, what is the maximum ruling spacing d? Ans: nm. How small can Δλ be if the 3rd order spectrum is used with this grating? Will the entire visible spectrum [400 to 700 nm] be seen in 3rd order for this grating? Explain how you know.

5 13.! Real light sources never produce only one frequency; there is always a spread of frequencies in the light. So the interference pattern of a film is never as sharp as the equations suggest. This fuzziness increases as the thickness of the film increases, until the pattern gets completely washed out. You are to show why.! Consider interference in light reflected at normal incidence from a thin air film of thickness t between parallel glass plates. The light consists equally of all frequencies between f and f + Δf. For frequency f the phase difference between the waves reflected from the top and bottom of the air film is δ 1 = 2mπ (m is an integer), so there is constructive interference in the reflected light for this frequency. For frequency f + Δf, the phase difference is δ 2 = (2m + 1)π, giving destructive interference for this frequency. Frequencies between f and f + Δf will give reflected intensities between these two extremes. For such a light source the net reflected intensity will be the same as if there were no interference at all. Write the formula for the phase difference between the waves in terms of the frequency f and the thickness t of the film. Find the value of Δf in terms of t for which the situation described occurs. Let the shortest and longest wavelengths in the light be 500 nm and 501 nm. For what value of t will the situation described occur? Ans: 62.5 µm.!![a film this thick or thicker will not give a discernible interference pattern for this source.]

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