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Unformatted text preview: PHYSICS 9B MIDTERM 2 December 1, 2009 I certify by my signature below that I will abide by the UC Davis Code of Academic Conduct. This includes • not copying from anyone else’s exam • not letting any other student copy from my exam • not discussing this exam with any student who has not yet taken it, nor providing any information, written or oral, that might get to a student who has not yet taken it. Anyone suspected of cheating will be automatically reported to Student Judicial Affairs.
LAST NAME: (P R I N T) FIRST NAME: (P R I N T) STUDENT ID: (LAST 4 DIGITS) Signature: Problem 1. (2 points) When light travels from medium “X” to medium “Y” as shown in the figure: A. both the speed and the frequency increase B. both the speed and the wavelength increase C. both the wavelength and the frequency are unchanged D. both the speed and the wavelength decrease E. both the speed and the frequency decrease Explanation: According to the figure, the index of refraction n of “X” is larger than that of “Y”. Therefore, both the speed and wavelength are increased by a factor of n . The frequency remains constant. Answer B. Problem 2. (2 points) If the path difference between two waves coming to a particular point is 4/3 wavelengths, the corresponding phase difference is Α. B. C. D. E. 3π/8 3π/4 4π/3 8π/3 0 X Y Explanation. One wavelength corresponds to the phase difference of 2π. Consequently 4/3 of the wavelength corresponds to 2π*4/3 = 8π/3. Answer D. Problem 3. (2 points) Which of the following represents a correct reflection of a ray from a shiny silvered Christmas tree ornament? A
reflected ray reflected ray reflected ray B C incident ray incident ray incident ray Solution. As this is an example of a convex mirror, the reflected ray appears to come from the focal point. Answer A. Problem 4. (2 points) An image of a coin embedded in the crystal ball is A. B. C. D. Real Erect Real Inverted. Virtual Erect. Virtual Inverted. coin Solution. The image is definitely virtual since the outgoing rays and the image are at the opposite sides of the sphere. To understand whether it is erect or inverted draw a ray that is immersed from the center of the coin to find the location of the image (see figure). Draw a second ray that is immersed from the top point of the coin to see that it will form an erect image. Answer C. Problem 5 (3 points) When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with film to cancel part of the glare. a) If the glass has a refractive index of 1.62 and you use a TiO2 coating with refractive index 2.62, what is the minimum film thickness that will cancel light of wavelength 505 nm? b) If this coating is too thin to stand up the wear, what other thicknesses would also work? b) If you decide to change the coating with refractive index 1.4 which is less than 1.62 what would the minimum film thickness be now? Solution. a) Condition of destructive interference when one of the two waves has a halfcycle reflection phase shift is given by 2t=mλTiO2 tmin=λTiO2 /2=(505/2.62)/2 =96.4 nm air 1.00 coat 2.62 glass 1.62 b) Any t equal to an integer multiplied by 96.4 nm will work as the coating thickness c) In case the refractive index of coating becomes less than 1.62, both reflective waves have a halfcycle phase shift, therefore relative phase shift is zero and the condition for destructive interference is 2t=(m+1/2) λcoating tmin=λcoating/4=(505/1.4)/4 =90.2 nm air 1.00 coat 1.4 glass 1.62 Problem 6 (3 points) Two convergent lenses with the same focal length f=1 cm are placed at the distance f from each other. The object is position at distance p=4/3f from the first lens. 1. Draw principle ray diagram determining the position of the image. 2. Find the position of the image with respect to lens 2 3. Find its lateral magnification. Is the image erect or inverted? Solution 1. Object f Image 2. We can for example solve this problem by applying thin lens equation twice. First find the position of the image produced by lens 1 using 1/p+1/q=1/f where all numbers are positive. We obtain 1/q=1/f1/p=13/4=1/4 and q=4 cm. Subsequently this object is located at distance equal to 3 cm with respect to the second lens and it is virtual object therefore it is p=3 cm that has to be used in the equation for the second lens: 1/p+1/q=1/f and 1/q=1/f1/p=1+1/3=4/3, hence q=3/4 cm. 3. The image is definitely inverted according to the picture. Now to prove this consider first the lateral magnification due to the first lens. It is given by m1=q/p=4/(4/3)=3 so that we obtain a factor of three larger image due to first lens that is inverted. Consider now this image as the object for the second lens which is virtual. The lateral magnification of it due to the second lens is m2=q/p=(3/4)/(3)=+1/4. The total lateral magnification is the product of both and is given by m1*m2=3*(1/4)=3/4. Problem 7 (3 points) Interference effects are produced at point P on a screen as a result of direct rays from a 431 nm source and reflected rays off the mirror, as shown. The source is 119 m to the left of the screen, and 0.52 cm above the mirror. a) What is the total number of bright fringes (those indicating complete constructive interference) seen at the screen above the mirror? b) At what angle relative to original direction of the beam will the bright fringe that is most distant from the horizontal plane of the mirror occur? c) If the maximum intensity is 2 W/m2, what is the intensity at points of the screen corresponding to the path difference of 2/3 wavelengths? Solution. Note that since the mirror has a higher refraction index than the air, the reflected light undergoes the phase change by half cycle therefore the conditions of the constructive and destructive interference interchange. That is dsinθ=mλ is the condition for destructive interference. dsinθ=(m+1/2)λ is the condition for constructive interference. In particular we see that the angle θ=0 (plane of the mirror) corresponds to the dark fringe! a) dsinθ=(m+1/2)λ therefore (m+1/2)λ/d=sinθ<1 λ/d==431x109/1.04x102= 0.0000414
mmax<d/λ−0.5=24129.930.5=24129.43 mmax=24129 b) Max. angle sinθmax =(mmax+1/2)λ/d=24129.93*0.0000414=0.9989 θmax=87 degrees
c) The path difference of 2/3 wavelengths corresponds to the phase difference φ=2π*2/3=4π/3. The intensity is given by I = I 0 cos2 φ / 2 = I 0 cos2 2π / 3 = I 0 / 4 = 0.5 W / m 2 Problem 8 (3 points) A twoslit Fraunhofer interferencediffraction pattern is observed with light of wavelength 640 nm. The slits have widths of 0.02 mm and are separated by 0.42 mm. 1. Draw the intensity as a function of angle that it will be seen on the screen. 2. What is the angular position of the central diffraction maximum? 3. How many bright fringes will be seen in the central diffraction maximum? Solution. 1. The intensity in this case will be a combination of diffraction and interference pictures, that behaves like this I(θ) θ 2. The condition for the first diffraction minimum is sinθ1=l/w so that the angular position of central diffraction maximum is given by sinθ1=l/w=640*109/0.02*103=0.032, hence θ1=0.032 rad=1.83 degrees and the central diffraction maximum spans from 1.83 to +1.83 degrees. 3. The condition for the mth interference maximum is sinθm=ml/d, hence if θm=θ1=0.032 rad , the mth interference maximum is given by m= sinθ1 *d/l=0.032*0.42*103/640*109=21. Notice however that the 21th maximum cannot be seen because it will be suppressed by the first diffraction minimum exactly at this angle. Therefore the last seen interference maximum corresponds to 20. If we count them from the left and from the right and include the central one, we obtain 41 as the number of bright fringes seen within central diffraction maximum. ...
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 Fall '09
 RichardScalettar
 Physics

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