Midterm Exam Solutions

Transcription

1 SIMG-455 Midterm Exam Solutions 1. We used the Argand diagram (also called the phasor diagram) to represent temporal oscillatory motion. (a) Use the Argand diagram to demonstrate that the superposition of three oscillations with identical frequencies is an oscillation with that frequency. An oscillation with angular temporal frequency ω 0 is represented on an Argand diagram as a vector with length equal to the amplitude of the oscillation that rotates around the origin at frequency ν 0 ω 0 Hertz. This is true of all such oscillations. The superposition of any number (not just three) oscillations with frequency ω 0 evaluated at t 0yields the vector sum of these oscillations. Since all of the component vectors rotate with frequency ν 0,somustthesum.Inshort, the sum of oscillations with frequency ν 0 yields an oscillation with frequency ν 0, regardless of the component amplitudes. (b) Use the Argand diagram to find the sum of these three oscillations: h f 1 [t] cos ν 0 t + i h f [t] cos ν 0 t i f [t] cos [ν 0 t] where ν 0 is arbitrary. Note that: cos [ν 0 t]cos[ν 0 t + ] PlottheseonanArganddiagramsothat: n h ³ f 1 [t 0] Re exp +i ν 0 0+ io h exp +i i 1 + i n h ³ f [t] Re exp +i ν 0 0 io h exp i i 1 i f [t] Re{exp [+i (ν 0 0 )]} exp[+i0] 1 + 0i 1

2 The sum of these at t 0is: Ã! 1 f 1 [t 0]+f [t 0]+f [t 0] + i + So the summation oscillation has amplitude 0. Ã 1 i! +( 1+i 0) 0

3 . A function f 1 [z,t] is created by summing two sinusoidal functions f 1 [z,t] cos z ν 0 t cos z ν 0 t Find an expression for the result as the product of two sinusoidal functions. Recall that: A + B A B cos A +cosb cos cos You can use this directly (if you remember it) or you can derive it as we did in class. I ll do that below for completeness: cos A +cosb Re e +ia + e +ibª n o Re e +i A e +i A + e +i B e +i B n ³ ³ o Re e +i A e +i A e +i B e i B + e +i B e +i B e +i A e i A because e +i A e i A e i 0 1. Now continue by rearranging: n ³ ³ o Re e +i A e +i A e +i B e i B + e +i B e +i B e +i A e i A n³ ³ ³ ³ o Re e +i A e +i B e +i A e i B + e +i A e +i B e i A e +i B n³ ³ o A+B A B A B +i +i i Re e e + e ½ ³ µ ¾ A+B +i A B Re e cos A B n o A+B +i cos Re e A B A + B cos cos So with that in hand, we can rewrite the sum of the two traveling-wave oscillations: cos z ν 0 t cos z ν 0 t µ cos z ν 0 t +cos zλ0 ν 0 t + µ µ cos z +ν 0 t + +cos z ν 0 t + µ cos z +ν 0 t + +cos z ν 0 t +

4 z λ cos 0 + z cos + ν 0t ν 0 t z z 4 cos z ν 0t ( ν 0 t) + cos [0 + ν 0 t +0] f 1 [z,t] 4 cos z + cos [ν 0 t] standing wave 4 sin zλ0 cos [ν 0 t] ifyouwanttogothisfar. 4

5 . Consider four ideal linear polarizers oriented at θ 1 0, θ 0, θ 60,and θ Natural light with intensity I 0 (proportional to E 0)isincidentonthefirst polarizer such that the emerging irradiance is I 1. (a) Specify the state of polarization of the light emerging from the last linear polarizer. Since the last element is a linear polarizer oriented at θ 4 90,thelightemerging (if any) MUST BE linearly polarized at θ 4 90 (b) Find the intensity or irradiance emerging from the last polarizer in terms of I 0 ; you may use any method, but extra credit will be given for use of Jones vectors and matrices. Using Jones vectors, the light emerging from the first polarizer has the form: E 1 I Recall that the Jones vector for light that is linearly polarized at azimuth angle θ is: cos θ E θ I sin θ The matrix for a polarizer at angle θ may be derived easily enough. We need two vector equations to evaluate the four unknown elements of the matrix. If the matrix represents a linear polarizer at angle θ, thenitmusttransmitthatjones vector: a b c d cos θ sin θ cos θ sin θ and block the Jones vector for light that is linearly polarized at angle θ + a b cos θ + c d sin a b sin θ 0 θ + c d cos θ 0 Combine into one matrix equation: µ µ a b cos θ sin θ c d sin θ cos θ µ µ cos θ 0 sin θ 0 a b cos θ sin θ c d sin θ cos θ cos θ 0 sin θ 0 Now solve for the matrix: a c b d a c a c cos θ 0 sin θ 0 b cos θ sin θ cos θ sin θ d sin θ cos θ sin θ cos θ b cos θ 0 cos θ sin θ d sin θ 0 sin θ cos θ cos θ sin θ sin θ cos θ 1 cos θ cos θ sin θ 1 cos θ sin θ sin θ 5

6 So the matrices for the different angles are: cos M 0 cos sin cos sin sin cos M 60 cos sin cos sin sin M 90 cos cos sin cos sin sin ³ ³ Now we can string the matrices together with the first one acting first: M M 90 M 60 M Now apply to the light emerging from the first polarizer: ME I Ã 16 I 0! 0 1 Linearly polarized light along the y direction with intensity 16 I 0 0.5I 0 6

7 4. :From observations of snowflakes, you might surmise that frozen water has a crystalline structure in a hexagonal form (due to the dominant angular position of the hydrogen atoms at 10 ). This anistotropic structure produces different forces on the charges in the molecules, which results in slightly different refractive indices along orthogonal directions. If the indices of refraction along orthogonal directions happen to be n and n 1.09, determine the thickness d of ice that would produce a relative phase delay of φ for the polarizations along the two directions. The phase delay through thickness d is determined by the number of wavelengths of light within the medium, which in turn is determined by and n : wavelength in medium with index n is λ n number of such waves in length d is N d λ nd number of radians in length d is Φ nd If two different indices n 1 and n, the phase differences will differ along the two different axes (i.e., everything is different). Φ 1 n1d Φ nd φ Φ 1 Φ d (n 1 n ) For a specified φ, we can rearrange the equation to find d: d φ Now substitute the known parameters: d 500 nm ( ) (n 1 n ) 500 nm 1, 50 nm 1.50 μm 0.01 mm 4( ) This is VERY thin, so something must be changed to create a useful quarter-wave plate. Exercise for the student. 7

8 5. Light is dispersed by a glass prism to create the spectrum as shown: (a) Replicate this sketch on your paper and label the order of the colors that emerge from the prism if the glass exhibits normal dispersion with a curve of this shape (vertical axis of n [λ] is arbitrary, but you may assume a reasonable range of values of n) The familiar ROYGBIV order of wavelengths; red is deviated the least and blue the most. (b) Now consider some theoretical substance that exhibits anomalous dispersion with a resonance centered at λ 500nm. Use the undamped model of the refractive index to sketch a possible graph of n [λ] for this material. One possible graph of the refractive index near the resonance is: 8

9 so the refractive index decreases up to the resonance and then jumps to a much larger value. (c) Make another sketch of a prism and label the sequence of dispersed colors that would be produced if the prism were made of the material in part (b). Explain any features on your graph. Onepossiblesolution: The index of refraction for blue light is now much smaller than it was before and the index for red light is now similar to that of blue. The largest index is for wavelengths just longer than 500 nm (i.e., green and yellow) 9