This relates to the frequency by: Then the result for C in terms of the given quantities is:

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1 . An AM rao station broadcasts at a frequency f = 830 khz. You receive that broadcast using a simple LC circuit which has an inductor L=85.0 mh and a variable capacitor. a) (8 points) You tune your car rao to achieve the best reception. What is the value of C? The resonance angular frequency of a LC circuit is given by: ω = / LC rad/s This relates to the frequency by: ω = π f Then the result for C in terms of the given quantities is: 3 C = /( ω L) = /(4 π f L) = F = 0.43pF b) (8 points) The amplitude of the electric field of the electromagnetic field that reaches your antenna is E ma = 0. V/m.What is the intensity of the electromagnetic wave reaching your antenna? We know the maimum value (amplitude) of E. The Poynting vector relates the average value of E (and B) to the intensity of the wave. For a sin wave, E average =E ma / and B average =B ma /, so we get an epression in terms of the amplitude (maimum value) of E. E ε ce ma 0 ma S = I = = =.33 0 W / m = 3.3 µ W / m µ c 0 5 c) (8 points) You are 3 km away from the rao transmitter. Assuming that the source transmits uniformly in all rections, what is the total power output of the transmitter? The transmitter emits uniformly in all rections. This means that spherical waves propagate through space. When the wave front reaches our antenna 3 km away from the source, we detect the intensity (which is power per unit area). Since we know the stance to the source, we can compute the total emitted power: 5 3 P = Intensity Area = I 4π r =.33 0 W / m 4 π (3 0 ) m = 504W d) (0 points) You are driving away from the transmitter on a road that has rect view to the transmitter (i.e. no reflections or absorption of the em waves). Your car rao circuit needs at least E ma = 0.0 V/m to be able to receive the signal. At what stance from the transmitter will the reception stop, if when you were 3 km away E ma was 0. V/m? We know the total emitted power from the source. As we move away from the source, the intensity of the wave will reduce as the area over which the total power is stributed becomes larger. The amplitude of the electric field of the received wave will also get smaller. At some stance it will become equal to E ma = 0.0 V/m. You will not be able to receive the signal at a stance which is larger than that. OK. Now for the calculation:

2 P = I 4 π(3 km) = I 4π I I 3km 3km (3 km) = Ema cε 0 I = E I (3 km) E = = 3km I3km E3km 0. V / m = 3km = 3km = 30km E 0.0 V / m Some constants that you may need: Speed of light : c = m/s Permittivity of free space: ε o = C / N.m Permeability of free space: µ 0 = 4π 0-7 T.m/A

3 . Shown in the figure below is a system containing an object and two lenses. Use the shape of the lenses in the figure to decide whether they are converging or verging optical elements. a) (8 points) Find the image location and magnification of the first lens assuming that only this lens eists. Specify the position of the image d i as some number of centimeters to the left or to the right of the first lens. do = 0cm f = cm + = do f 0 = = 30 cm ( right of lens) 8 30 m = = =.5 d0 0 b) (8 points) The image from the first lens is the object for the second one. Find the image location and magnification of the second lens. Specify the position of the image d i as some number of centimeters to the left or to the right of the second lens. The image from the first lens is to the right of the second one. Thus the object for the second lens is NOT on the incoming side of the light. The lens is verging, so f is also negative.

4 d o = cm f = 40cm + = d 40 i ( 40) ( ) = = cm ( right of lens) m = =. c) (5 points) Find the magnification of the total system m= mim =.5i. = 3.33 d) (5 points) Is the image real or virtual? Is it upright or inverted? Real, inverted. e) (0 points) You want to produce an image at the same location as the result above using a single lens placed mid-way between the two lenses. Find the focal length of this lens. What is the magnification in this case? You need to put one single lens ( remove the two lenses which form a zoom lens) at 4 cm from the object and produce an image at = 5.89 cm. = + f f = 6.5cm The new magnification is: 5.89 m = =. 4

5 3. You have two apertures and a laser with wavelength λ = 600 nm. You shine the laser through the first aperture. Figure below shows the light intensity measured on a screen that is located.5 m behind the aperture. Then you shine the laser through the second aperture (removing the first one) and the intensity on the screen (.5 m away) is shown in Figure. Figure Figure a) Is the aperture from Figure a single slit or a double slit? If the aperture is a single slit, what is its width? If it is a double slit, what is the spacing between the slits? Two closely spaced slits produce a double-slit interference pattern with the intensity maima that are equally spaced and equal width. If the slit has finite width, the ffraction causes the maima to decrease (slowly) in amplitude as we go off the central maimum. The structure due to the ffraction is much broader than the interference structure, since the slit width is ALWAYS smaller than the slit spacing. In Figure our screen catches several intensity maima that are due to the double-slit interference. These are all within the first broad maimum which is caused by the ffraction. We are asked to just find the slit spacing, thus we measure the stance between the interference maima ( cm) and then get: dsinθ = mλ m =, θ tanθ sin θ = λ/ d = y/ L 9 λ (600 0 m)(.5 m) L d = = = 0.5 mm y 0.00 m Note: I used the small angle approimation, but you will get (almost) the same result, if you find the tan and sin eactly. b) Is the aperture from Figure a single slit or a double slit? If the aperture is a single slit, what is its width? If it is a double slit, what is the spacing between the slits? OK, here we have a central maimum that is double the width of the side maima. The light intensity decreases rapidly when we go off the central maimum. Thus, Figure corresponds to a single slit aperture. From the figure the separation between the central maimum and the first minimum is y =.0 cm =.0 0 m. Now, in ffraction, we have a contion for the MINIMA given in terms of the slit width:

6 asinθ = mλ m =, θ sinθ tan θ = λ/ a = y/ L (.5 m)( m) a = Lλ / y = = 0.5 mm.0 0 m c) You have another laser with unknown wavelength. You shine the laser through the first aperture ( Figure ) and you find that the m = 4 bright fringe from the unknown laser overlaps the m = 3 bright fringe from the λ = 600 nm laser. What is the unknown wavelength? We have double slit interference. The bright fringes are located at positions given by dsinθ m = mλ. For the m = 3 bright fringe from the know-wavelength laser, the interference contion is dsinθ ( 9 3 = m). For the m = 4 bright fringe from the unknown-wavelength laser the contion is dsinθ 4 = 4λ. Because the position of the fringes is the same, 9 9 ( ) ( ) 3 dsinθ = dsinθ = 4λ = m λ = m = 450 nm 3 4 4