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1 Exam 3 Solutions Prof. Paul Avery Prof. Pradeep Kumar Apr. 6, Four charges are placed along a straight line each separated by a distance L from its neighbor. The order of the charges is +Q, Q, Q, +Q. What is the total potential energy of the system (relative to infinity)? Answer: 5kQ / 3L Solution: Let the charges be ordered 1,, 3, 4. There are 6 combinations: (1, 3, 4), (13, 4), 14 which give contributions kq / L 5kQ / 3L for the total potential energy. ( ) + ( +kq / L) + ( kq / L) + ( +kq / 3L) =. A wire of length L has a resistance of 60Ω. The wire is uniformly stretched to five times its original length, maintaining constant volume. If a length L is now cut from the stretched wire, what is the resistance of the remaining piece? Answer: 100Ω Solution: Stretching the wire at constant volume gives 5 times the length and 1/5 the cross sectional area, giving a total resistance of = 1500Ω. After the piece is cut off, the resistance of the remaining wire is 4/5 of the new resistance or = 100Ω. 3. In the circuit shown in the figure, what is the potential difference V a - V b between points a and b? Answer: +1.0 V b Solution: This is two-loop problem. Let the top loop have a 4Ω 4Ω clockwise current i 1 and the bottom loop have a clockwise current i. Thus the current going through b is i 1 + i moving to the left. The equations describing the two loops are 16 4i 1 +1i 18 = 0 (top) and 18 1i 8 i 1 + i ( ) = 0 (bottom). Solving yields i 1 = 1.0 and i = 0.5, so the current flowing to the left through point b is 1.5 A. Starting from point a the voltage falls by = 1 volts, thus V a V b = +1.0 volts. There are other ways of defining the loops (e.g., defining i as the current flowing through point b to the left), but the current moving through point b to the left is always 1.5 A. 16 V + 18 V 1Ω + 4Ω a 1

2 4. Positive point charges q and 4q are located at x = 0 and x = 4d, respectively. What is the magnitude of the electric field at x = d? Answer: 5kq / 9d Solution: For positive q, the charge at x = 0 generates an E field of kq / d to the right while the other charge gives an E field of 4kq / 9d in the opposite direction, so the magnitude of the net field is 5kq / 9d. 5. Three point charges are placed as shown at the corners of a right triangle (L =.0 cm, q = 4.0 µc). What is the y component of the force on the +q charge? Answer: +39 N +q L -q L +q Solution: The y component of the force from the q charge is kq / 4L. The corresponding y force component from the +q charge is ( kq / 5L )sinθ, where sinθ = / 5. The total y force component is thus ( kq / L )( 4 5 / 5 1/ 4) = +39 N.

4 10. The figure shows a mass spectrometer in which charged particles of different masses enter a region through a slit and move perpendicular to a uniform magnetic field permeating the region (the gray dots represent the magnetic field pointing out of the page). For each particle, the detector measures the distance between the entrance point to the place where it strikes the bottom of the region. Assume 1 C and 14 N singly charged ions are accelerated to the same velocity before entering the spectrometer. If the 1 C ions strike the detector 3.75 m from the slit, at one point will the 14 N ions hit it? Answer: 4.38 m Solution: From the equation equating centripetal force to the magnetic force, mv / R = qvb, we find R = mv / qb. Thus for the same velocity ions will strike the surface at a distance proportional to the mass. The distance for 14 N ions is then d = ( 14 / 1)3.75 = 4.38m. 11. The intensity of the sunlight reaching the Earth's surface is approximately 700 W/m. What is the rms value of the electric field in V/m? Answer: Solution: The total intensity is related to the rms electric field by u = ε 0 E rms E rms = 513 V/m.. Solving yields 1. Unpolarized light is incident on four polarizing sheets with their transmission axes oriented relative to the horizontal by angles of 15, 60, 15 and 60, respectively. What percentage of the original intensity is transmitted? Answer: 6.3 Solution: The first polarizer reduces the intensity by 50%. All subsequent polarizers are oriented 45 relative to one another so for each one cos θ = 1. Thus the total fraction of light transmitted is 1 ( 1 ) 3 = 1 16 ~ 6.3%. 13. A light ray in the core (n = 1.40) of a cylindrical optical fiber is incident on the cladding surrounding the core. The ray is transmitted through the cladding (n = 1.5) and then into the air. The emerging ray makes an angle 0.0 with the outside surface of the cladding. What angle did the ray in the core make with the normal? Answer: 4. Solution: This is solved by Snell s law using n core sinθ core = n air sinθ air, with θ core = 70. Solving yields θ air = 4. relative to the normal. 4

5 14. A woman sits 40 cm in front of a makeup mirror. Her upright image is magnified by 1.5 times. What is the focal length of the mirror? Answer: +10 cm Solution: The magnification is M = q / p = 1.5, thus q = 60 cm. Using the lens equation with p = 40 yields f = +10 cm. 15. An inverted image is formed 50 cm from an object by a thin lens located between the two. The image is 1/ the height of the object. What is the distance from the object to the lens? Answer: 33 cm Solution: The magnification is M = q / p = 0.5, and p + q = 50 cm. Thus q = 1 p. Solving yields p = 50 / 1.5 = 33 cm. 16. You are given a microscope with an objective focal length f o = 10 cm and an eyepiece of focal length f e = 0 cm. If you want to focus the microscope on a specimen cm away from the objective lens, how far apart must the objective and eyepiece be from each other? Answer: 45 cm Solution: Applying the lens equation to the objective lens gives q = +5 cm. This image is located at the focal point of the eyepiece. Thus the total length of the tube is = 45 cm. 17. Refer to the previous question. If you were to take the lenses out of the microscope and make a telescope (maximum power) out of them, what will be the tube length? Answer: 30 cm Solution: The maximum angular magnification is m θ = f e / f o =. Thus the tube length is = 30 cm. We also accepted None of these because of potential confusion with the definition of tube length for a microscope. 18. In a double slit experiment, the intensity of the light reaching the center of he screen from one slit alone is I 0 and intensity of the light reaching the center from the other slit alone is 9I 0. When both slits are open, what is the intensity of the light at the minima nearest the center? Answer: 4I 0 Solution: The electric field at the minimum is the difference of the electric fields. The intensity ( ) = 4I 0. is proportional to the square of the electric field, so the net intensity is 9I 0 I 0 5

6 19. A thin film of soapy water (n = 1.37) is held vertical. Light of wavelength 600 nm is reflected from the film and horizontal dark lines appear. What is the thinnest soap film for which a bright band will occur? Answer: 109 nm Solution: For a soap bubble, the condition for an interference maximum is d = ( m + 1 )λ, where λ = 600 / 1.37 = 438 nm is the wavelength in the bubble. Thus the thinnest value of d is 438/4 or 109 nm. 0. If the two nd order maxima (m = ) are separated by.0 cm on the screen in a double-slit experiment, what is the separation of the m = 3 minima? Assume the angle is very small. Answer: 3.5 cm Solution: The condition for a double slit maximum is d sinθ = mλ, which for small angle is θ ~ mλ / d. The minimum is located at θ ~ ( m + 1 )λ / d. Thus the m = 3 minimum is located 1.75 times the distance to the second order maximum (m = ) or 3.5 cm. 6

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