practice-final7-solution

Question 1: In a new quantum system, a transition between levels 2 and 3 is caused by a photon being absorbed with a wavelength of ? ? . A transition between levels 3 and 4 is caused by a photon being absorbed with a wavelength of 2 3 ? ? . a) (2 pts) In this system, is the energy gap between levels getting smaller, larger, or staying the same as you go to higher levels. Explain your answer. Since the wavelength is getting smaller, the frequency and thus the energy is getting larger with higher levels. b) (5 pts) Calculate the wavelength emitted with a transition in this quantum system occurs from level 4 to 2, ? 42 . You should express your final answer in terms of ? ? . Show your work. Δ? 42 = Δ? 43 + Δ? 32 hc λ 42 = hc λ 43 + hc λ 32 = hc 2 3 ? 0 + hc λ 0 = 5 2 ℎ? ? 0 ? 42 = 2 5 ? 0 c) (3 pts) You have two identical double-slit experimental setups, with a slit separation of 0.02 mm. You send the photons emitted from the 3 to 2 transition in this quantum system through the 1 st experiment, and photons emitted from the 4 to 3 transitions through the 2 nd double-slit setup. You adjust the distance to the screen in the 2 nd setup such that the two interference patterns are identical. If the distance to the screen is 80cm in the 1 st setup, what is the distance in the second setup? You may use the small angle approximation. Show your work. Small angle approximation: Δ𝑦 = ?𝜆𝐷 ? Set-up 1: Δ𝑦 = ?𝜆 0 𝐷 1 ? Set-up 2: Δ𝑦 = ? 2 3 𝜆 0 𝐷 2 ? Setting them equal to each other, ?𝜆 0 𝐷 1 ? = ? 2 3 𝜆 0 𝐷 2 ? , and solving for D 2 results in: ? 2 = 3 2 ? 1 = 3 2 80cm = 120 cm

Questions 2: a) (5 pts) When we described amateur optometry in 7C, we neglected the distance between the glasses and the eye. Thus, the prescription we found was more appropriate for contact lenses, which have direct contact with the eye, than for glasses which sit some distance in front of the eye. Assume Kate needs corrective lenses for both reading and seeing distant objects. Her contact lens prescriptions for both conditions are -2D and 1.5D. Using the same principles of amateur optometry, calculate Kate ’s glasses prescription if the distance between the glasses and the lens of the eye is 3 cm. The diameter of the lenses in the glasses is 15 mm. Recall, the near point of healthy eyes is 25 cm as measured from the eye. You might find it helpful to draw a schematic. Show your work. For nearsightedness, we want the image to be at the far point of the patient and the lens to be diverging. For contact lenses as we learned in DL, the focal length of the corrective lens, ? ? is just the far point, fp, of the patient: 1 ? 𝑙 = 1 ∞ − 1 fp , so f l = −fp . For Kate f l = − 1 2m = −0.5 m However, when there is a distance, ? , between the eye and glasses, the image formed by the glasses needs only to be a distance fp − d away from the glasses to be fp distance away from the eye. Thus, with glasses the image distance of the virtual image formed is i = −(fp − ?). Thus, the focal length of glasses, ? ? is: 1 ? 𝑔 = 1 ∞ − 1 fp−d , so f g = −(fp − d) = f l + d Writing in terms of optical power: ? ? = 1 op g = 1 op l + d = −0.5m + 0.03m = −0.47 m . Thus, the optical power of her glasses is: op g = − 1 0.47 m = −2.13 D Similarly for farsightedness, the image formed by the glasses with a converging lens needs to be a distance of np − d from the glasses, such as that image is a distance equal to the near point, ?? , away from the patient’s eye : i = −(np − ?). First calculate Kate’s near point, given that the object is at 25 cm=0.25 m: 1 ? 𝑙 = 1 ? − 1 np ; and 1 ?? = 1 ? − 1 ? 𝑙 = 1 0.25m − 1.5 m . Resulting in: np = 0.4 m Thus, for glasses i = −(np − d) = −(0.4 m − 0.03) = −0.37 m. The object distance is also closer to the glasses by 0.03, ? = .25 − 0.03 = 0.22m 1 ? ? = op g = 1 ? + 1 𝑖 = 1 0.22m − 1 0.37m = 1.84D b) (5 pts) Iris has transition glasses (2 different lenses) which correct both her nearsightedness and farsightedness. The optical power of the two lenses are 2D and -1.5D. She takes off her glasses and places them in front of a candle. She measures the magnification by the converging lens to be double the magnification by the diverging lens. Both images are virtual. Find the distance between the candle and the glasses. Show your work. Let the first lens be the converging one: op 1 = 1 ? 1 = 2D and ? 1 = − 𝑖 1 ? Let the second lens be the diverging one: op 2 = 1 ? 2 = −1.5D and M 2 = − 𝑖 2 ? = 𝑀 1 2 . Relating image distances, we get: 𝑖 2 = − 1 2 ? 1 ? = − 1 2 (− 𝑖 1 ? ) ? = 𝑖 1 2 For lens 1: op 1 = 1 ? + 1 𝑖 1 . Therefore, for lens 2: op 2 = 1 ? + 1 𝑖 2 = 1 ? + 2 𝑖 1 Plugging in: 1 𝑖 1 = op 1 − 1 ? for lens 1 into the equation for lens 2: op 2 = 1 ? + 2 (op 1 − 1 ? ) = − 1 ? + 2op 1 Solving for object distance: 1 o = 2op 1 − op 2 = 2 × 2D − (−1.5D) = 5.5D, resulting in ? = 1 5.5 m = 0.182 m = 18.2 cm

Questions 4: Shown below is an electron which is traveling out of the page at 2x10 6 m/s. Also shown are equipotential lines, generated by some source charged material, which make 20 o with the horizontal. At this instance of time the electron is also in between two current carrying wires as shown. The wire on the left has current of 5A flowing into the page, while the direction and magnitude of the current in the other wire is unknown. The total magnetic field due to the two wires at the location of the electron is 1.875x10 -7 T pointing down. The resistance of the wires per meter is 0.05 Ω ? . Some relevant distances are marked in the diagram. a) (3pt) Find the electric force exerted on the electron. Express your answer in terms of magnitude and angle. Clearly draw the vector marking the angle you determined. Show your work. |? ⃗ | = Δ? ?? = 0.2? 0.3 ? = 2 3 ? ? |? | = |?||? ⃗ | = 1.6 × 10 −19 ? × 2 3 ? ? = 1.07 × 10 −19 ? Force points in the direction of increasing voltage for a negative charge, resulting in 70° Northeast. b) (3pt) Determine the direction and magnitude of current in wire 2. Show your work. ? ⃗ ??? = −1.875 × 10 −7 T Using RHR#1 the magnetic field due to wire 1 points down at the location of the electron: ? ⃗ 1 = − ? 0 𝐼 1 2𝜋? 1 = 4π × 10 −7 N A 2 × 5A 2𝜋 × 0.7 m = −1.43 × 10 −6 T Since the magnitude of ? ⃗ 1 is bigger than that of ? ⃗ ??? , ? ⃗ 2 needs to point up. So current is into the page. ? ⃗ 2 = ? ⃗ ??? − ? ⃗ 1 = −1.875 × 10 −7 T + 1.43 × 10 −6 T = 1.24 × 10 −6 T = ? ? 𝐼 2 2𝜋? 1 I 2 = |? ⃗ 2 |2πr 1 ? 0 = (1.24 × 10 −6 T)(2π)(0.5 m) 4π × 10 −7 N A 2 = 3.1 A c) (4pt) Calculate the total force acting on the electron. Express your answer in terms of magnitude and angle. Clearly draw the vector marking the angle that you calculated. Show your work. Electric force: ? ? = −1.07 × 10 −19 Ncos 70° = −3.66 × 10 −20 N ? ? = 1.07 × 10 −19 N sin 70° = 1.0 × 10 −19 N ? ? = (−3.66 × 10 −20 N, 1.0 × 10 −19 N) Magnetic force points in the -x direction based on RHR#2: |? | ? = |?||? ⃗ ??? ||𝑣 | = 1.6 × 10 −19 C × 1.875 × 10 −7 T × 2 × 10 6 m s = 6 × 10 −20 N ? ? = (−6 × 10 −20 N, 0) ? ??? = ? ? + ? ? = (−9.66 × 10 −20 N, 1.0 × 10 −19 N) |? | ??? = √(9.66 × 10 −20 ) 2 + (1.0 × 10 −19 ) 2 = 1.39 × 10 −19 ? 𝜃 = tan −1 ( 1.0×10 −19 9.66×10 −20 ) = 46° pointing northwest.

Questions 5: The top view of two coils of wire is shown below. The side view shown the outer coils. The inner loop is made up of 1,000 turns of copper wire with a radius of 𝑅 𝑖???? = 20?? . The outer loop is made up of 10,000 turns of the same copper wire with a radius 𝑅 ????? = 25?? . A controllable current runs in the outer wire loops as shown in the top view. a) (3pt) If current in the outer loops is decreased over time, is there an induced current in the inner loops? If so, in what direction does it flow and what direction is the magnetic field generated by this induced current as viewed from the top view? (Clearly define all directions and show all work). As the current is decreased the external magnetic field produced by the outer coil of wires, into the page by RHR1, also decreases. If we chose our normal to point out of the page in the top view, the flux decreases with decreasing field leading to an induced magnetic field also into the page. This field must be produced by a clockwise current as seen from the top view. b) (5pt) We can approximate the magnitude of the magnetic field inside a stack of circular loops as ? 0 ?𝐼/? , where L is the length of the loop stack (see side view) and N is the number of turns. What is the necessary change in current to produce an average EMF of 24V in 5ms inside the inner loop stack? Both the inner and outer stacks have a length of 50cm. Show your work. The EMF is given by ℰ = − ? 𝑖???? ∆Φ Δ? , which can be expanded to find the change in current. ℰ = − N inner ∆Φ Δt = − N inner ∆( BA) Δt = − N inner × A inner μ 0 N outter L × ΔI Δt Solving for the change in current yields Δ𝐼 = |ℰ |LΔt N inner N outter π r inner 2 μ 0 = (24V ) (0 .5m ) (0 .005s ) ( 1000)( 10 ,000)( π 0 .2 2 m 2 ) ( 4π × 10 −7 N A 2 ) = 0 .038A c) (2pt) A permanent magnet is then inserted into the inner wire stack. Will the outer loop stack need a smaller, larger, or the same change in current to induce the same EMF as in part b) on the inner loop stack? No calculations are needed, but fully explain and justify your answers. The magnet acts to change the strength of the magnetic field inside both current loops. However, it does not change when the current is changed, so it does not contribute to the change in magnetic flux. Therefore, the change in current to induce an EMF of 24V will be the same as found in part b).