Physics Revision Exercises PDF

1 EXERCISES FOR REVISION OF PHYSICS EXAM FOR ADVANCED LEVEL  VERIFY THE ANSWER PROPOSED BEFORE TRYING YOURSELF I WISH YOU ALL THE BEST  2 3. Explain briefly the role of machines in agriculture in rapid development of the count...

1 EXERCISES FOR REVISION OF PHYSICS EXAM FOR ADVANCED LEVEL  VERIFY THE ANSWER PROPOSED BEFORE TRYING YOURSELF I WISH YOU ALL THE BEST  2 3. Explain briefly the role of machines in agriculture in rapid development of the country towards suitable programs of transformation and modernization of agriculture. (8 marks) SOLUTION They are used to till the ground (1 mark), plant seed and perform other tasks (1 mark). -Application of machineries in the day to day farm activities increases marginal output in food production (1 mark) and poverty eradication (1 mark). Sustainable agricultural mechanization can also contribute significantly to the development of value chains (1 mark) and food systems as it has the potential to render postharvest, processing and marketing activities (0.5 mark) and functions more efficient (0.5 mark), effective and environmentally friendly. Sustainable mechanization can: - increase land productivity by facilitating timeliness and quality of cultivation (1 mark); - support opportunities that relieve the burden of labour shortages (1 mark) and enable households to withstand shocks better; - decrease the environmental footprint of agriculture when combined with adequate conservation agriculture practices; and - reduce poverty and achieve food security while improving people's livelihoods. 4. Plan and write a short note on the uses of Optical fibers in different fields with clear explanation. (10 marks) SOLUTION Consider only five(5) correct uses. Optical fiber is made up of thin strands of glass used to transmit information and data in the form of light. The fields that utilize optical fibers are different; 1. In Medical industry (1 mark) Because of its extremely thin and flexible nature, it is used in various instruments to view internal body parts (0.5 mark) by inserting into hollow spaces in the body. It is used as lasers during surgeries, endoscopy, microscopy and biomedical research. The advent of practicable optical fibers has seen the development of much medical technology. Optical fibers have paved the way for a whole new field of surgery, called laparoscopic surgery (or more commonly, keyhole surgery), which is usually used for operations in the stomach area such as appendectomies. By using an optical fiber to carry the laser beam to the relevant spot (0.5 mark) which would then be able to be used to cut the tissue or affect it in some other way.  3 2. In the communication system (1 mark) The telecommunication has major uses of optical fiber cables for transmitting and receiving purposes. It is used in various networking fields and even increases the speed and accuracy of the transmission data. Compared to copper wires, fiber optics cables are lighter, more flexible and carry more data (the greatest bandwidth) (1 mark). 3. In defense purpose (1 mark) Fiber optics is used for data transmission in high-level data security fields of military and aerospace applications. These are used in wirings in aircraft, hydrophones for SONARs and Seismic applications (1 mark) 4. In Industries (1 mark) These fibers are used for imaging in hard-to-reach places such as they are used for safety measures and lighting purposes in automobiles both in the interior and exterior. They transmit information at lightning speed and are used in airbags and traction control. They are also used for research and testing purposes in industries. (1 mark) 5. Optical Fibers used for broadcasting (1 mark) These cables are used to transmit high-definition television signals which have greater bandwidth and speed. Optical Fibre is cheaper compared to the same quantity of copper wires. Broadcasting companies use optical fibres for wiring HDTV, CATV, video-on-demand and many applications. (1 mark) 6. Uses of Optical Fiber for Lightening and Decorations (1 mark) By now, we got a fair idea of what is optical fiber and it also gives an attractive, economical and easy way to illuminate the area and that is why it is widely used in decorations and Christmas trees. (1 mark) 7. Optical fibers used in Mechanical Inspections (1 mark) On-site inspection engineers use optical fibers to detect damages and faults which are at hard-to-reach places. Even plumbers use optical fibers for the inspection of pipes. (1 mark)  4 5. In Predicting the future of the universe, Could you predict how will the universe look after billion and billion years? (6 marks) Answers Telescopes allow us to study our surroundings, to see the cosmos and understand the laws of physics that gave way to the development of multicellular life forms. Telescopes are also time machines. Through them, we can look back through the history of our universe and see some of the very first celestial objects that were birthed from the Big Bang. But telescopes allow us to see far more than our past. With them, we see our future. We can determine the rate at which the universe is expanding, see stars be born and die in equal proportions, detect changes in the atmosphere of distant exoplanets, and so much more. It has been said that, due to the accelerating expansion of the universe, the sky we're observing today will look radically different from the one that'll exist in a few trillion (or even billion) years. So, assuming the universe exists in a state similar to how it is now — without a big rip, big freeze, big bounce, big slurp, or any other cosmos-ending scenarios taking place — what will our descendants see when they look out into the vast reaches of the cosmos? Or rather, what will they not see? If the size of the universe were to remain unchanged, the mutual gravitational attraction among galaxies eventually would cause all of them to merge together. But as we‘ve known ever since the astronomer Edwin Hubble‘s 1929 discovery, the universe is expanding and galaxies, on average, are moving farther apart. Throughout much of the 20th century, the big question in cosmology circles was: Is there sufficient mass in the universe to enable gravity to halt this expansion? Or will distant galaxies continue to move apart, slowing down but never quite stopping? By conclusion, In the 1920s, astronomer Edwin Hubble discovered the universe was not static. Rather, it was expanding; a find that revealed the universe was apparently born in a Big Bang. After that, it was long thought the gravity of matter in the universe was certain to slow the expansion of the universe. Then, in 1998, the Hubble Space Telescope's observations of very distant supernovae revealed that a long time ago, the universe was expanding more slowly than it is today. In other words, the expansion of the universe was not slowing due to gravity, but instead inexplicably was accelerating. The name for the unknown force driving this accelerating expansion is dark energy, and it remains one of the greatest mysteries in science.  5 Hence, The Universe will expand forever: If the mean density is less than the critical density, then there is insufficient mass within the universe to stop the expansion - the universe will expand forever. Ultimately, the galaxies will move increasingly further apart. 6. Radioactivity is the spontaneous emission of radiation from the nucleus of an unstable atom. Elements which exhibit radioactivity are said to be radioactive. Radioactivity is the spontaneous and random emission of radioactive rays from unstable radioactive materials after which they become more stable. Propose the uses of radioactivity in different field with clear explanations in your own words. ( 10 marks) Solution Consider only five (5) uses (0ne use carries 1 mark and clear explanation carries 1 mark) 5=10 marks Application of radioactivity Industrial uses: 1. In food manufacturing industries and preservation, gamma rays are used in sterilizing food such as meat so that it can stay fresh for a longtime. 2. Radioisotopes are used in detecting leakages in underground pipes carrying water or oil. A little radioactive solution is added to the liquid being pumped at the source. The point of leakage is found by monitoring the soil surrounding the pipe. Presence of a leakage creates temporary radioactivity in the soil surrounding the leak and this can be detected. 3. Beta particles in the paper industry are used to control the thickness of paper automatically during production process. This is done by having a beta source below the paper and a GMT with a counter above it. Any change in thickness of paper produces a corresponding change in the number of counts detected by the counter. 4. Determining the rate of the wear in piston rings. A piston ring is irradiated so that it becomes radioactive. The mass of the piston ring and its initial activity are determined. The ring is then installed in the engine oil and the engine is run continuously for some time. After this time, the activity of the oil is then determined using a GMT. This activity is due to mass of the piston worn off. 5. Medical uses Gamma rays are replacing X-rays in the treatment of cancer. I know it sound ironic that the very radiation which causes cancer, can also be used to treat it. This is of course possible by use of modern techniques. Radiopharmaceuticals are used for medical imaging to scan the internal tissues (hard tissues) Radioisotopes are used as tracers for diseases like brain tumors. Sterilizing medical equipments like needles, syringes, surgical knives before packaging.  6 Radioactivity can be used to assess the blood volume of a patient. 6. Agricultural uses Radioactive elements are used to measure the rate of uptakes and distribution of fertilizers, within plants. Radioactive substances can be used to sterilize the pests in the plants. 7. Archaeological uses Radioisotopes mainly carbon-14 are used to study the age of fossils and rocks. This process is called carbon dating. 7. Explain how Radio Tuning works with Capacitance and Inductance (4 marks) a) How could you explain the relationship of the capacitance to frequency? (2 marks) Solution a) What the variable capacitor and the inductive coil are doing is making a circuit that is resonant at one frequency. (1 mark) All the other frequencies are attenuated away while the resonant frequency is passed through. (1 mark) A capacitor and an inductor in parallel, as in a common crystal radio, form a band-pass filter. The pair resonates at a certain frequency (determined by the values of inductance and capacitance) (1 mark). At this resonant frequency, current will circulate between the two components in near-perfect balance, and the pair (in parallel) will present an extremely high impedance, looking almost like an open circuit, i.e., as if they weren't there at all. The current can then proceed through the detector and headphones. At frequencies that are very different from the resonant frequency, the pair will look almost like a short circuit(1 mark). Specifically, at very low frequencies, the inductor will look like not much more than a piece of wire, and at very high frequencies, the capacitor will act that way. Because of this, no current will be available for the detector and headphones. OR A particular station is ―tuned in‖ by adjusting C and or L so that the resonant frequency of the circuit equals that of the station‘s carrier frequency. When a coil and capacitor are said to be tuned to resonance, the inductive and capacitive reactances are equal but opposite in action. When this condition is met, the reactances cancel one another. The tuned circuit then looks like a pure resistance at the frequency of resonance, a desirable condition. b) The relationship of the capacitance to frequency is: the higher the capacitance, the lower the frequency (1 mark). This is because a higher value capacitor requires longer to charge than a lower value capacitor. (1 mark)  7 b) 30. Explain how Radio Tuning works with Capacitance and Inductance (4 marks) c) How could you explain the relationship of the capacitance to frequency? (2 marks) Solution c) What the variable capacitor and the inductive coil are doing is making a circuit that is resonant at one frequency. (1 mark) All the other frequencies are attenuated away while the resonant frequency is passed through. (1 mark) A capacitor and an inductor in parallel, as in a common crystal radio, form a band-pass filter. The pair resonates at a certain frequency (determined by the values of inductance and capacitance) (1 mark). At this resonant frequency, current will circulate between the two components in near-perfect balance, and the pair (in parallel) will present an extremely high impedance, looking almost like an open circuit, i.e., as if they weren't there at all. The current can then proceed through the detector and headphones. At frequencies that are very different from the resonant frequency, the pair will look almost like a short circuit(1 mark). Specifically, at very low frequencies, the inductor will look like not much more than a piece of wire, and at very high frequencies, the capacitor will act that way. Because of this, no current will be available for the detector and headphones. OR A particular station is ―tuned in‖ by adjusting C and or L so that the resonant frequency of the circuit equals that of the station‘s carrier frequency. When a coil and capacitor are said to be tuned to resonance, the inductive and capacitive reactances are equal but opposite in action. When this condition is met, the reactances cancel one another. The tuned circuit then looks like a pure resistance at the frequency of resonance, a desirable condition. d) The relationship of the capacitance to frequency is: the higher the capacitance, the lower the frequency (1 mark). This is because a higher value capacitor requires longer to charge than a lower value capacitor. (1 mark) 8. Q1. i. Slice of tissue x-ray is (1 mark) A. Tomography B. Mammography C. Contrast studies D. All of above Answer: A  8 ii. The basic principles behind ultrasound probe technology is (1 mark) A. Piezoelectric effect B. Photoelectric effect C. Calorimetric effect D. Raman effect Ans: A. Piezoelectric effect iii. Radiation produced by nuclear decay / disintegration (1 mark) A. Gamma rays B. X-ray C. Proton beam D. Cosmic rays Ans: A. Gamma rays iv. Ionizing radiation is the substance which ionizes the atoms when it passes. It is harmful for biological matter. Which of the following is non-ionizing modality? (1 mark) A. Conversational x-ray B. Computerized tomography C. Magnetic resonance imaging D. Isotopic scanning Ans: C. Magnetic Resonance Imaging Q2. a) What does MRI stand for? (1 mark) b) What is the purpose of MRI? (2 marks) c) Does MRI use Ionizing radiation? (2 marks) ANSWER: a) Magnetic Resonance Imaging. (1 mark) b) It is a mode of medical imaging which produces pictures of the body‘s anatomy (1 mark) and physiology. (1 mark) c) Magnetic resonance imaging is a type of medical imaging that uses radio waves and a magnetic field (1 mark). MRI does not use ionizing radiation. (1 mark) Q3. i. Choose the correct answer. Risks associated with radionuclide imaging are: (1 mark)  9 a. Generally poor resolution compared with other imaging modalities. b. Rarely receiving an overdose of chemical injected in the vein of the body. c. High capital and running costs d. None of them. Answer: d. None of them ii. Choose the correct answer. The medical imaging techniques used sounds are: (1 mark) a. Radionuclide b. Radiography c. Mammography d. Endoscopy Answer: c. Mammography 9. Q1. a) What facts explain that Mammography films show breast calcifications better than chest x-rays because? (3 marks) b) Which one of type of an X-Ray radiation Mammography uses for? (1 mark) Answer a) -Makes a more uniform exposure of the entire breast (1 mark) -Reduces geometric unsharpness (1 mark) -Reduces exposure time and scattered radiation (1 mark) b) Mammography is specialized medical imaging that uses a low-dose x-ray (1 mark) system to see inside the breasts. Q2. i. Which statement is correct? Answer by True if it is correct and by False if it is not correct. a. The use of gamma radiation to form images following the injection of various radiopharmaceuticals is known as Scintigraphy. (1 mark) b. This decision to scan or not to scan a normal pregnancy must be made only by the photographer. There are universally accepted guidelines at present. (1 mark) c. Tissue in the body absorbs and scatters ultrasound in the same ways. Lower frequencies are more rapidly absorbed (attenuated) than higher frequencies. (1 mark) d. Upper endoscopy uses light and camera to view the esophagus, stomach, and upper part of the small intestine. (1 mark)  10 e. Ultrasound is both generated and detected through high frequency oscillations in piezoelectric crystals so there is ionizing radiation exposure associated with ultrasound imaging. (1 mark) ii. How does a pregnancy care provider perform an abdominal ultrasound? What is the purpose of this exam? (2 marks) Answer i. a. True b. False c. False d. True e. False 1 mark for each ii. A pregnancy care provider performs an abdominal ultrasound by placing a transducer (ultrasound probe) directly on skin (0.5 marks). Then, they move the transducer around your belly (abdomen) (0.5 marks) to capture images of your baby (1 mark). a) 33. Explain why a gel is used between the skin and the transducer when an ultrasound scan of a fetus is taken. (2 marks) b) For ultrasound of frequency 3.5 MHz, the acoustic impedance of muscle is 1.78 × 106 kg m−2 s−1, and that of soft tissue is 1.63 × 106 kg m−2 s−1. (2 marks) ANSWER a) There is a large difference between the acoustic impedance of air and skin. ( 0.5 marks) Consequently, a very large percentage of the ultrasound is reflected( 0.5 mark). The gel is used to match the impedances or to prevent the deviation of ultrasound. (1mark) b) (2 marks) 10. Analyze the figure below and answer the questions  11 a) What is the type of medical imaging technique shown above? (1 mark) b) Explain how it works? (5 marks) c) Outline the reasons why for scanning during a normal pregnancy (3 marks) d) Is this imaging technique an operator dependent rather than other? Why? (2 marks) Solution a) Echography exam or Ultrasonic imaging technique or ultrasound system (1 mark) b) The process of imaging is the same as the echo-locating sonar of a submarine or a bat (0.5 marks). The observer sends out a brief pulse of ultrasound and waits for an echo (0.5 marks). The pulse travels out from the ultrasound prob (0.5 marks), reflects off the target and returns. The ultrasound machine uses pulses because the same device acts as both transmitter and receiver (0.5 marks). The emitted ultrasound waves are reflected back to the crystal by the various tissues of the body (0.5 marks). These reflected sound waves also called the “echoes” act on the piezoelectric crystal in the ultrasound probe to produce an electric signal (0.5 marks), again by the piezoelectric effect (0.5 marks). It is this electric signal which is analyzed by a computer produces a cross- sectional image (0.5 marks). c) Reasons for scanning during a normal pregnancy are:  12 To see age of foetus (1 mark), position of foetus (1 mark), development of foetus (1 mark) etc d) Yes, It is highly operator dependent (1 mark) as it relies on the operator to produce and interpret images at the time of examination(1 mark). 11. Q1.i.What is radiation? A. The emission of light from a source B. The transfer of heat through conduction C. The emission and transmission of energy in the form of electromagnetic waves or particles D. The release of gases from a chemical reaction Answer C. The emission and transmission of energy in the form of electromagnetic waves or particles ii. What is the unit used to measure radiation exposure? A. Radians B. Sieverts C. Coulombs D. Grays Answer D. Grays iii. Which of the following is a common application of ionizing radiation in medicine? A. MRI scans B. Ultrasound imaging C. X-ray imaging D. Thermography Answer C. X-ray imaging iv. What is the main source of background radiation? A. Nuclear power plants  13 B. Medical imaging devices C. Cosmic rays, rocks, and soil D. Cell phone towers Answer C. Cosmic rays, rocks, and soil Q2.a) What is meant by radiation absorbed dose? Give its SI unit. (2 marks) Answer a) Radiation absorbed dose is a measure of the energy of deposited in a medium by ionizing radiation(1 mark). In the SI system of units, the unit of measure is joules per kilogram, and its special name is gray (Gy) (1 mark). b)Malignant neoplasms (cancer), hereditary mutations, cellular death and skin burnt are the effects of ionizing radiation. (2 marks) Which one mentioned above are example(s) of Deterministic effects or stochastic effects? Answer Deterministic effects: cellular death and skin burnt (1 mark) Stochastic effects: Malignant neoplasms (cancer), hereditary mutations (1 mark) 12. Q1.. How can I ensure that personnel who work in my lab, but do not use radioactive material, do not violate the security requirements? (4 marks) Solution: The security requirements apply to the material, not the people. Therefore, all personnel in your lab must be trained (2 marks) and cognizant of the security requirements for the material in the lab they are working in. (2 marks) Q2. Differentiate Ionizing radiation and Non-ionizing radiations (4 marks) Give two examples for each. Answer This is a radiation that carries enough energy to liberate electrons from atoms or molecules, thereby ionizing them (2 marks). As the more powerful form of radiation, ionizing radiation is more likely to damage tissue than non-ionizing radiation. The  14 main source of exposure to ionizing radiation is the radiation used during medical exams such as X-ray radiography or computed tomography scans. (2 marks). Non-ionizing radiation refers to any type of electromagnetic radiation that does not carry enough energy to ionize atoms or molecules (2 marks).. Examples of non- ionizing radiations include visible light, microwaves, ultraviolet (UV) radiation, infrared radiation, radio waves, radar waves, mobile phone signals and wireless internet connections. (2 marks). APPLYING How much energy is absorbed when a person of mass 70 kg receives an effective dose equivalent to 30 mSv, half the dose equivalent being acquired from radiation of quality factor 1 and half from radiation of quality factor 3. (5 marks) Solution H1=D1 1= 15 mSV (1 mark) D1= 15 mJkg-1 (1 mark) H2=D2 = 15 mSV (1 mark) D2=5 mJkg-1 (1 mark) (1 mark) 13. Imagine you want to observe one method of evaluating atomic crystalline structure by using x-ray diffraction and to understand the use of Bragg's Law and its relation to crystal structure a) Can you propose the X-Ray Diffraction Instrumentation Description? (3 marks) Explain the function of each apparatus. (8 marks) Answer Apparatus (2 marks), each clear explanation carries 2 marks The X-ray diffraction experiment requires an X-ray source (0.5 mark), the sample under investigation (0.5 mark) and a detector or photographic film (0.5 mark) to pick up the diffracted X-rays and filter or collimator (0.5 mark), a monochromator, a recorder I. X-ray Source The X-ray source is a circuit composed of an anode, a cathode, and a battery. The electric current from the battery heats the cathode which is made up of a tungsten filament, that  15 emits thermionic electrons. The emitted electrons hit the anode which has the target metal, transferring the energy to the surface of the target that produces X-rays. The target material selected should have an atomic number greater than that of the elements being examined in the sample. The energy of x-rays emitted by the target material should be greater than that required to excite the elements being irradiated. (2 marks) II. Collimator The target material produces the x-rays which are always randomly directed. They are converted into a narrow beam of X-rays by a collimator. The collimator consists of two metal plates separated by a small gap. It absorbs all the X-rays except the narrow beam that passes through the gap. (2 marks) III. Monochromators There is a wide range of wavelengths possessed by the x-rays but only a short range is useful for the analysis. Monochromators help to pass only the X-rays with desirable wavelengths. There are usually two types of monochromators used in X-ray diffraction instruments: Filters Filters act as a window of material that absorbs undesirable radiations but allows the radiation of the required wavelength to pass. This method makes use of the large difference in the mass absorption coefficient on the other side of an absorption edge. For example, a zirconium filter is used for molybdenum radiation. (1 mark) Crystal Monochromators A crystal monochromator is positioned in the path of the X-ray beam so that the angle reflecting planes satisfy Bragg‘s equation for the required wavelength. The beam is split up by the crystalline material into the component wavelengths in the same way as a prism splits up white light into rainbow colors. This is called analyzing crystals. They can be flat crystal or curved crystal monochromators made up of lithium fluoride, sodium chloride, or quartz.(1 mark) IV. Detectors The detectors are placed after the monochromators to detect the signals and transfer them to the recorder. (2 marks) b) The figure below represents the Reflection of x-rays from two planes of atoms in a solid.  16 From the Braggs‘s Law Interactive, you observed how to derive Bragg‘s Law: n  2d sin  Explain (in words) why n cannot be a half-integer value Answer: n is an integer, it is the order of maxima to have constructive interference. Constructive interference occurs only when the path length difference between rays scattered from parallel crystal planes would be an integer number of wavelengths of the radiation. When the crystal planes are separated by a distance d, the path length difference would be 2dsin θ. 14.  17 Wheatstone bridge counts two conditions: 1. When the bridge is balanced, the current through the galvanometer is zero, IG=0 2. I3=Ix and I1=I2  18 Example: Consider the circuit below Find Rx. 19  20 15. Using the K.e. theorem, find the acceleration of the following system: A worker lifts up a stone of 3.5kg to a height of 1.80m each 30s. Find the work done in one hour.  21 16. We actually know fifteen satellites revolving around the planet Uranus. Let us denote the period of revolution of satellite by T and the mean distance to the centre of the planet by r. The five bigger than others have the following characteristics: a) (i) For each satellite, calculate T2 and r3, (ii) Assume T2 =y and r3 = x. Trace the graph of y = f(x). What conclusion related to the nature of the graph can you get? b) (i) Calculate the slope of the plotted segment. (ii) Deduce the mass of Uranus 17. Spacecraft at 𝒓𝑬, What is the force of gravity acting on a 2000kg spacecraft when it orbits two Earth radii from the earth‘s center (that is, a distance 𝑅𝐸 = 6380𝑘𝑚 above the earth‘s surface, the mass of the Earth is 𝑀𝐸 = 5.98 × 1024𝑘𝑔. 18. A geosynchronous satellite is one that stays above the same point on the earth, which is possible only if it is above a point on the equator. Such satellites are used for TV and radio transmission, for weather forecasting, and as communication relays. Determine (a) the height above the Earth‘s surface such a satellite must  22 orbit, and (b) such a satellite‘s speed. (c) Compare to the speed of a satellite orbiting 200km above Earth‘s surface. OTHER EXERCISES OF S4 1. How many electrons are contained in 1.0C of charge? What is the mass of the electrons in 1.0C of charge? Answer: 6.2 1018 electrons, 5.7 1012 kg 2. If two equal charges, each of 1 C, were separated in air by a distance of 1km, what would be the force between them? Answer: 9kN repulsion 3. What is the force of repulsion between two argon nuclei that are separated by 1.0nm? The charge on an argon nucleus is 18e. Answer: 75nN 4. Determine the force between two free electrons spaced 1.0 angstrom apart. Ans. 23nN repulsion. 5. Two equally charged balls are 3cm apart in air and repel each other with a force of 40 N. Compute the charge on each ball. Ans.: 2nC. 6. Three point charges are placed at the following points on the x-axis: 2.0C at x  0,  3.0C at x  40cm,  5.0 C at x  120cm. Find the force (a) on the 3.0 C charge, (b) on the 5.0 C charge. Ans.: (a) 0.55N ; (b) 0.15N 7. Two charged metal plates in vacuum are 15cm apart as shown in figure below. Electric field between the plates is uniform and has a strength of E  3000 N / C. An electron ( q  e, me  9.11031 kg ) is released from rest at point at P just outside the negative plate. (a) How long will it take to reach the other plate? (b) How fast will it be going just before it hits?  23 8 Ans.: (a) 2.4 10 seconds , (b) 1.30 10 m / s 7 8. Suppose in figure above (in 16th question) an electron is shot straight upward from point P with a speed of 5.0 106 m / s. How far above A will it strike the positive plate? Ans. 12cm above point A. 9. In figure above, a proton (q  e, m  1.67 1027 kg ) is shot with speed 2.00 105 m / s toward P from A. What will be its speed just before hitting the plate at P? Ans.: 356km/s. 10. Two identical tiny metal balls have charges q1 and q2. The repulsive force one exerts on the other when they are 20cm apart is 1.35 104 N. After the balls are touched together and then separated once again to 20cm, the repulsive force is found to be 1.406 104 N. Find q1 and q2. Ans.: 20nC, 30nC. Alternatively, both charges could have been negative. 11. Three charges are placed on three corners of a square, as shown below. Each side of the square is 30.0cm. Compute the electric field at the fourth corner. What would be the force on 6.00 C charge placed at the vacant corner?  24 Ans. : 2.47 105 N / C at 118o , 1.48 N at 118o. 12. Four equal-magnitude point charges (3.0C ) are placed at the corners of a square that is 40cm on a side. Two, diagonally opposite each other, are positive, and the other two are negative. Find the force on either negative charge. Ans.: 0.46N inward along the diagonal. 13. Determine the acceleration of a proton (q  e, m  1.67 1027 kg ) in an electric field of strength 0.50kN / C. How many times is this acceleration greater than that due to gravity? Ans.: 4.8 1010 m / s 2 , 4.9 109 14. A 0.200-g ball hangs from a thread in a vertical electric field of 3.00kN / C directed upward. What is the charge on the ball if the tension in the thread is (a) zero and (b) 4.00mN? Ans.: (a) +653nC; (b) -680nC 15. A tiny, 0.60-g ball carries a charge of magnitude 8.0  C. It is suspended by a thread in a downward 300N/C electric field. What is the tension in the thread if the charge on the ball is (a) positive, (b) negative? Ans.: (a) 8.3mN; (b) 3.5mN 16. As shown in figure below, a charged particle remains stationary between the two horizontal charged plates. The plate separation is 2.0cm , and m  4.0 1013 kg and q  2.4 1018 C for the particle. Find the potential difference between the plates.  25 Solution Since the particle is in equilibrium, the weight of the particle is equal to the upward electrical force. That is, mg  qE or mg  4.0 10 kg  9.81m / s  13 2 E  18  1.63 106V / m q 2.4 10 C But for a parallel-plate system, V  Ed  1.63  106V / m   0.020m   33kV. 17. An alpha particle  q  2e, m  6.7 1027 kg  falls from rest through a potential drop of 3.0 10 V  3.0MV . (a) What is its KE in electron volts? (b) 6 What is its speed? Solution qV  2e   3.0 10   6.0 10 eV  6.0MeV 6 (a) Energy in eV   6 e e 1 2 1 2 (b) PE E  KE gained  qV  mv f  mvi 2 2  2  1.6 1019 C  3.0 106V    6.7 1027 kg  v 2f  0 from which v f  1.7 107 m / s. 1 2 18. Two metal plates are attached to the two terminals of a 1.50-V battery. How much work is required to carry a 5.0C charge (a) from the negative to the positive plate, (b) from the positive to the negative plate? Ans.: (a) 7.5 J , (b) 7.5 J 19. The plates described in problem above are in vacuum. An electron  q  e, me  9.11031 kg  is released at the negative plate and falls freely to the  26 positive plate. How fast is it going just before it strikes the plate? Ans.: 7.3 105 m / s 20. A proton  q  e, m p  1.67 1027 kg  is accelerated from rest through a potential difference of 1.0MV. What is its final speed? Ans.: 1.4 107 m / s 21. An electron gun shoots electron  q  e, me  9.11031 kg  at a metal plate that is 4.0mm away in vacuum. The plate is 5.0V lower in potential than the gun. How fast must the electrons be moving as they leave the gun if they are to reach the plate? Ans.: 1.3 106 m / s 22. The potential difference between two large parallel metal plates is 120V. The plate separation is 3.0mm. Find the electric field between the plates. Ans.: 40kV / m toward negative plate An electron  q  e, me  9.110 kg  is shot with speed 5.0 106 m / s parallel 31 23. to a uniform electric field of strength 3.0kV / m. How far will the electron go before it stops? Ans.: 2.4cm 24. A potential difference of 24kV maintains a downward-directed electric field between two horizontal parallel plates separated by 1.8cm. Find the charge on an oil droplet of mass 2.2 1013 kg that remains stationary in the field between the plates. Ans.: 1.6 1018 C  10e 25. Determine the absolute potential in air at a distance of 3.0cm from a point charge of 500 C. Ans.: 15kV 26. Briefly describe how a lightening conductor can safeguard a tall building from being struck by lightning. 27. Find the force between two point charges +4μC and -3μC placed at a distance of 12dm apart in free space. 28. A charge of 4μC is placed in a vacuum. Determine the electric field 29. Intensity at a point P at a distance of 20cm from the charge. 30. The vertical deflecting plates in a Television set are 5.0cm and 1.0cm apart. If a potential difference of 100V is applied between the plates and the electron beam enters horizontally mid-way between the plates with a speed of 2.0 × 10-7ms-1. Find the kinetic energy gained from the electric field by an electron in the beam. SOLVED QUESTIONS  27 31. Two coins lie 1.5m apart on a table. They carry identical charges. Approximately how large is the charge on each if a coin experiences a force of 2N? Solution The diameter of a coin is small compared to the 1.5m separation. We may therefore approximate the coins as point charges. Coulomb‘s law, k q1q2 FE  , gives (with K approximated as 1.00 in air) K r2 F r2 (2 N )(1.5m) 2 q1q2  q 2  E   5 1010 C 2 from which q  2 105 C. k 9 109 N.m2 / C 2 32. Repeat the problem (32) above if the coins are separated by a distance of 1.5m in a large vat of water. The dielectric of water is about 80. Solution k q2 From Coulomb‘s law, FE  where K , the dielectric constant, is now 80. Then K r2 FE r 2 K (2 N )(1.5m) 2 (80) q   2 104 C k 9 10 Nm / C 9 2 2 33. A helium nucleus has charge +2e, and a neon nucleus 10e, where e is the quantum of charge, 1.60 1019 C. Find the repulsive force exerted on one by the other when they are 3.0 nanometers apart. Assume the system to be in vacuum. Solution Nuclei have radii of order 1015 m. We can assume them to be point charges in this 19 qq ' 2 (2)(10)(1.6 10 C) case. Then FE  k 2  (9.0 10 N.m / C ) 9 2 9  5.11010 N  0.51nN r (3.0 10 m 34. Three point charges are placed on the x-axis as shown below. Find the net force on the 5C charge due to the other charges. Solution Because unlike charges attract, the forces on the 5C charge are as shown in figure. The magnitudes of F E3 and F E8 are given by Coulomb‘s law:  28 (3.0 106 C )(5.0 106 C ) FE3  (9.0 109 N.m2 / C 2 )  3.4 N (0.20m) 2 (8.0 106 C )(5.0 106 C ) FE8  (9.0 10 N.m / C ) 9 2 2  4.0 N (0.30m) 2 Notice two things about the computation: (1) Proper units (Coulombs and meters) must be used. (2) Because we want only the magnitudes of the forces, we do not carry along the signs of the charges. (That is, we use their absolute values.) The direction of each force is given by the diagram, which we drew from inspection of the situation. From the diagram, the resultant force on the center charge is FE  FE8  FE3  4.0 N  3.4 N  0.6 N and it is in the +x-direction. 35. Find the ratio of the Coulomb electric force to the gravitational force between electrons in vacuum. Solution From Coulomb‘s Law and Newton‘s Law of gravitation, q2 m2 FE  k and FG  G r2 r2 FE kq 2 / r 2 kq 2 (9.0 109 N.m 2 / C 2 )(1.6 1019 C ) 2 Therefore    11 31  4.2 1042 2 FG Gm / r 2 Gm 2 (6.67 10 N.m / kg )(9.110 kg ) 2 2 2 As you can see, the electric force is much stronger than the gravitational force. 36. As shown in figure below, two identical balls, each of mass 0.10g, carry identical charges and are suspended by two threads of equal length. At equilibrium they position themselves as. Find the charge on either ball.  29 Consider the ball on the left. It is in equilibrium under three forces: (1) the tension FT in the thread; (2) the force of gravity, mg  (1.0 104 kg )(9.81m / s 2 )  9.8 104 N And (3) the Coulomb repulsion FE. Writing F x  0 and  Fy  0 for the ball on the left, we obtain FT cos 60o  FE and FT sin 60o  mg  0 From the second equation, mg 9.8 104 N FT  o   1.13 103 N sin 60 0.866 Substituting in the first equation gives FE  FT cos 60o  (1.13 103 N )(0.50)  5.7 104 N But this is the Coulomb force, kqq '/ r 2. Therefore, FE r 2 (5.7 104 N )(0.40m) 2 qq '  q 2   from which q  0.10C. k 9.0 109 N.m2 / C 2 37. The charges shown in figure below are stationary. Find the force on the 4.0  C charge due to the other two. Solution From Coulomb‘s Law we have  30 6 6 qq ' 2 (2.0  10 C )(4.0  10 C ) FE2  k  (9.0  109 N.m 2 / C )  1.8 N r2 (0.20m)2 6 6 qq ' 2 (3.0  10 C )(4.0 10 C ) FE3  k 2  (9.0 10 N.m / C ) 9 2  2.7 N r (0.20m)2 The resultant force on the 4  C charge has components FEx  FE2 cos 60o  FE3 cos 60o  (1.8  2.7)(0.50) N  0.45 N FEy  FE2 sin 60o  FE3 sin 60o  (1.8  2.7)(0.866) N  3.9 N So FE  FE2x  FE2y  (0.45)2  (3.9) 2 N  3.9 N The resultant makes an angle of tan 1 (0.45 / 3.9)  7o with the positive y-axis, that is,   97 o. 38. For the situation shown below, find (a) the electric field E at point P, (b) the force on a 4.0 108 C charge placed at P, and (c) where in the region the electric field would be zero (in the absence of the 4.0 108 C charge). Solution (a) A positive test charge placed at P will be parallel to the right by the positive 9.0 109 N.m2 / C 2 charge q1 and q2. Hence, E  2 (25 108 C )  9.0 105 N / C directed (0.050m) toward the right. (b) A charge q placed at P will experience a force Eq. Therefore, FE  Eq  (9.0 105 N / C )(4.0 108 C )  0.036 N. The negative sign tells us the force is directed toward the left. This is correct because the electric field represents the force on a positive charge. The force on a negative charge is opposite in direction to the field. (c) By reasoning, we conclude that the field will be zero somewhere to the right of the 5.0 108 C charge. Represent the distance to that point from the 5.0 108 C charge by d. at that point, E1  E2  0 ; because the field to the positive charge is to the right, while the field due to the negative charge is to the left.  31  q1 q2  2  20 10 C 8 5.0 108 C  Thus k  2  2   (9.0 10 N.m / C )   0 9 2  (d  0.10m) 2  r1 r2  d2  Simplifying, we obtain 3d  0.2d  0.01  0 which gives d  0.10m and  0.03m. 2 Only the plus sign has meaning here, and therefore d  0.10m. The point in question is 10cm to the right of the negative charge. Two charges are placed on the x-axis: 3.0C at x  0 and  5.0 C at x  40cm. Where must a third charge q be placed if the force it experiences is to be zero? Solution The situation shown in figure below: We know that q must be placed somewhere on the x-axis. (Why?) Suppose that q is positive. When it is placed in interval BC, the two forces on it are in the same direction and cannot cancel. When it is placed to the right of C, the attractive force from the 5C charge is always larger than the repulsion of the 3.0 C charge. Therefore, the force on q cannot be zero in this region. Only in the region to the left of B can cancellation occur. (Can you shown that this is also true if q is negative?) For q placed as shown, when the net force on it is zero, we have F3  F5 and so, for distances in meters, q(3.0 106 C ) q(5.0 106 C ) k  k d2 (0.40m  d ) 2 After canceling q, k, and 106 C from each side, we cross-multiply to obtain 5d 2  3.0(0.40  d )2 or d 2 1.2d  0.24  0 b  b 2  4ac 1.2  1.44  0.96 Using the quadratic formula, we find d    0.60  0.775m 2a 2 Two values, 1.4m and -0.18m, are therefore found for d. the first is the correct one; the second gives the point in BC where the two forces have the same magnitude but do not cancel. 39. In the figure below, the potential difference between the metal plates is 40 V. (a) Which plate is at the higher potential? (b) How much work must be  32 done to carry a 3.0C charge from B to A? From A to B? (c) How do we know that electric field is in the direction indicated? (d) If the plate separation is 5.0mm, what is the magnitude of the electric field? Solution (a) A positive test charge between the plates is repelled by A and attracted by B. Left to itself, the positive test charge will move from A to B, and so A is at the higher potential. (b) The magnitude of the work done is carrying a charge q through a potential difference V is qV. Thus the magnitude of the work done in the present situation is W  (3.0C )(40V )  0.12kJ Because a positive charge between the plates is repelled by A, positive work (+120J) must be done to drag the 3.0C charge from B to A. to restrain the charge as it moves from A to B, negative work (120 J ) is done. (c) A positive test-charge between the plates experiences a force directed from A to B and this is, by definition, the direction of the field. (d) For parallel plates, V  Ed. Therefore, V 40V E   8.0kV / m. d 0.0050m Notice that the SI units for electric field, V/m and N/C, are identical.  33 40. How much work is required to carry an electron from the positive terminal of 12-V battery to the negative terminal? Solution Going from the positive to the negative terminal, one passes through a potential drop. In this case it is V  12V. The W  qV  (1.6 1019 C )((12)  1.9 1018 J As a check, we notice that an electron, if left to itself, will move from negative to positive because it is a negative charge. Hence positive work must be don to carry it in the reverse direction as required here. 41. How much electrical potential energy does a proton lose as it falls through a potential drop of 5kV ? Solution The proton carries a positive charge. It will therefore move from regions of high potential to regions of low potential if left free to do so. Its change in potential energy as it moves through a potential difference V is V Vq. In our V  5kV. Therefore, case, Change in PE E  Vq  ( 5 103V )(1.6 1016 C )  8 10 16 J 42. An electron starts from rest and falls through a potential rise of 80V. What is its final speed? Solution Positive charges tend to fall through potential drops; negative charges, such as electrons, tend to fall through potential rises. Change in PEE  Vq  (80V )(1.6 1019 C )  1.28 1017 J. This lost PE E appears as KE of the electron: PE E  KE gained 1 1 1 (1.28 1017 J )(2) 1.28 1017 J  mv 2f  mvi2  mv 2f  0 from which v f  31  5.3 106 m / s 2 2 2 9.110 kg 43. (a) What is the absolute potential at each of the following distances from a charge of 2.0  C : r  10cm and r  50cm ? How much work is required to carry a 0.05 C charges from the point at r  50cm to that at r  10cm ?  34 Solution q 2.0 106 C 10 (a) V10  k  (9.0 109 N.m2 / C 2 )  1.8 105V ; V50  V10  36kV r 0.10 50 (b) Work  q(V10  V50 )  (5 108 C )(1.44 105V )  7.2mJ 44. Suppose, in problem (a), that a proton is released at r  10cm. How fast will it be moving as it passes a point at r  50cm ? Solution As the proton moves from one point to the other, there is a potential drop of Potential drop  1.80 105V  0.36 105  1.44 105V The proton acquires KE as it falls through this potential drop: KE gained  PE E lost 1 2 1 2 1 mv f  mvi  qV  (1.6 1027 kg )v 2f  0  (1.6 1019 C )(1.44 105V ) 2 2 2 from which v f  5.3 106 m / s. 45. A tin nucleus has a charge 50e. (a) Find the absolute potential V at a radius of 1.0 1012 m from the nucleus. (b) If a proton is released from this point, how fast will it be moving when it is 1.0m from the nucleus? Solution q (50)(1.6 1019 C ) (a) V  k  (9.0 109 N.m2 / C 2 )  72kV r 1012 m (b) The proton is repelled by the nucleus and flies out to infinity. The absolute potential at a point is the potential difference between the point in question and infinity. Hence there is a potential drop of 72kV as the proton flies to infinity. Usually we would simply assume that 1.0m is far enough from the nucleus to consider it to be at infinity. But, as a check, let us compute V at r  1.0m :  35 q (50)(1.6 1019 C ) V1m  k  (9.0 109 N.m2 / C 2 )  7.2 108V which is essentially r 1.0m zero in comparison with 72kV. As the proton falls through 72kV, KE gained  PE E lost 1 2 1 2 1 mv f  mvi  qV  (1.67 1027 kg )v 2f  0  (1.6 1019 C )(72000V ) 2 2 2 from which v f  3.7 106 m / s. 46. The following point charges are placed on the x-axis: 2.0C at x  20cm,  3.0 C at x  30cm,  4.0 C at x  40cm. Find the absolute potential on the axis at x  0. Solution Potential is a scalar, and so q  2.0 106 C 3.0 106 C 4.0 106 C  V  k  i  (9.0 109 N.m2 / C 2 )     ri  0.20m 0.30m 0.40m    9.0 109 N.m2 / C 2 10 106 C / m  10 106 C / m  10 10 6 C / m   90kV 47. Two point charges,  q and  q, are separated by a distance d. Where, besides at infinity, is the absolute potential zero? Solution q q At the point (or points) in question, 0  k k or r1  r2 r1 r2 This condition holds everywhere on a plane which is the perpendicular bisector of the line joining the two charges. Therefore the absolute potential is zero everywhere on that plane. 48. In figure below, the charge at A is 200 pC , while the charge at B is 100 pC. (a) Find the absolute potential at points C and D. (b) How much work must be done to transfer a charge of 500  C from point C to point D?  36 Solution  2.00 1010 C 1.00 1010 C    9.0 109 N.m2 / C 2   qi a) VC  k     2.25V ri  0.80m 0.20m   2.00 1010 C 1.00 1010  VD   9.0 109 N.m2 / C 2      7.88V  7.9V  0.20m 0.80m  b) There is a potential rise from C to D of V  VD  VC  7.88V  (2.25V )  10.13V. So W  Vq  10.13V   5.00 104 C   5.1mJ 11.7. End unit questions 1. Distinguish between isothermal and adiabatic changes, clearly stating the conditions under which they occur in practice. 2.. Define the two principal molar heat capacities of a gas and derive an expression relating the two. Explain the difference between these two principal molar heat capacities. 3. (a) What is meant by internal energy of a system? (b) Cubical room 2.5 x10 x7m3 is filled with nitrogen gas at 100kPa and 270C. What is the mass of gas in the room? c) How many calories are needed to increase the room temperature 10C if all the heat goes into the gas? (i) Assume constant volume and (ii) Assume constant pressure. 4. a) Explain why refrigerator acts as heat engine in reverse. b)How many kilograms of water at 0C can a freezer with a coefficient of performance 5 make into ice cubes at O0C with a work input of 3.6MJ(one kilowatt-hour) 5. A gas has a volume of 0.02 m3 at a pressure of 2 x 105 Pa and a temperature of 27o C. It is heated at constant pressure until its volume increases to 0.03 m3. Calculate the: (a) External work done. (b) New temperature of the gas. (iii) Increase in internal energy of the gas if its mass is 16g, its molar heat capacity at constant volume is 0.8Jmol-1K-1 and the molar mass is 32g. 6. An ideal gas at 17o C has a pressure of 760mm Hg is compressed (i) isothermally (ii) a diabatically, until its volume is halved. Calculate in each case the final temperature and pressure of the gas. Assume that Cp = 2100Jmol-1K-1 and CV = 1500Jmol-1K-1. 7. A quantity of oxygen is compressed isothermally until its pressure is doubled, it is then allowed to expand adiabatically until its original volume is restored. Find  37 the final pressure in terms of its original pressure. Draw a PV diagram for the above processes. 8. 0.45m3 of a gas at a temperature of 15o C expands adiabatically and its temperature falls to 4o C. a) What is the new volume if γ = 1.40 b) The gas is then compressed isothermally until the pressure returns to its original value. Calculate the final volume of the gas. 9.. A vessel containing 2m3 of air initially at a temperature 25o C and pressure 760mmHg, is heated at constant pressure until its volume is doubled. Find (a) the final temperature (b) the external work done by the air in expanding, (c) the quantity of heat supplied. (Assume that the density of air at s.t.p is 1.293kgm-3 and that the principal molar heat capacity of air at constant volume is 20.4Jmol-1K-1. 10. An ideal gas at a temperature 45o C and pressure 1.0 x 105Nm-2 occupies a volume of 2.0 x 10-3 m3. It expands adiabatically to twice its volume. Find the final temperature and pressure. Represent this process on PV- diagram. (Take γ = 1.40) 11. (a) (i) What is meant by a reversible isothermal change? (ii) State the conditions for achieving a reversible isothermal change. (b)(i) What is meant by adiabatic change? (ii) An ideal gas at 27oC and a pressure of 1.01x105 Pa is compressed reversibly and isothermally until its volume is heated. It is then expanded reversibly and adiabatically to twice its original volume. Calculate the final pressure and temperature of the gas if γ = 1.40. 12. (a) Explain why the specific heat of a gas at constant pressure is higher than that at constant volume. (b)The density of an ideal gas is 1.6kgm-3 at 27oC and 1.00 × 10Nm2 pressure and specific heat capacity at constant volume is 0.312KJkg-1. Find the ratio of the specific heat capacity at constant pressure to that at constant volume. Point out any significance attached to the result. 13. (a) Explain why the cooling compartment of a refrigerator is always on top. (b) The refrigerator cools substances by evaporation of a volatile liquid. Explain how evaporation causes cooling. (c) State the reason why water is used in the cooling system of a car engine. 14. (a) With the aid of a labelled diagram, describe how a refrigerator works. b)The cooling system of a refrigerator extracts 0.7 Kw of heat. How long will it convert 500g of water at 20oC to ice?  38 (c) Explain how evaporation takes place in the refrigerator. (d) Explain why water in a porous pot keeps at a lower temperature than that of the surrounding. 15. a) The equation of state of an ideal gas is PV=nRT. For each of these symbols, state the physical quantity and SI unit. Use a table like the one below. Symbol Physical quantity SI unit P V n R T a) 0.80moles of an ideal gas monoatomic is enclosed in a cylinder by a frictionless, perfect –fitting piston. The conditions are such that the gas expands at constant pressure (process AB) , it is then compressed at constant temperature (process BC)until its volume returns to original value. These changes are represented together with the relevant numerical data on the graph in the figure below. Analyze the figure above and answer the following questions (i) What is the pressure and the volume of the gas at point A? (ii) Which parts of the diagram correspond to isobaric and isochoric processes respectively? (iii) Show that the temperature of the gas at A is 301K (iv) What is the temperature of the gas at point B?  39 (v) Calculate the pressure of gas at point C. (vi) Calculate the work done during process AB (vii) Calculate the work done during process BC (viii) Calculate the work done during process CA (ix) Calculate the work output during the cycle. 16. a) With aid of three elements of consideration differentiate Gasoline to Diesel engine b) Give two impacts of heat engine on climate change. c) For a gasoline engine undergoing the Otto cycle shown in figure below, Show that the thermal efficiency of a gasoline engine operating in an Idealized Otto cycle is given by  1 V  i.   1   2  and  V1  T T ii.   1  A  1  D TB TC 12.8. End unit questions I. Choose the most suitable answer from the options 1.The angular distance of an object around the horizon, starting from the north, and measured eastwards around the horizon to a point on the horizon directly below the object‘s location on the celestial sphere is known as the: (a) Horizon (b) Latitude (c) Longitude (d) Altitude (e) Azimuth 1. The angular distance above the celestial horizon is called the: (a) Horizon (b) Latitude (c) Longitude (d) Altitude  40 (e) Azimuth 2. This is a point in the sky that‘s located directly above the observer: a. Horizon b. Latitude c. Longitude d. Azimuth e. Zenith Extended questions 3. Describe and investigate in your group the Astronomical scale 4. Work in group and investigate the Sun-Earth-Moon system: (eclipses, and phases of the moon) a) Solar eclipse and Lunar eclipse b) Phase of the Moon 5. In your group describe the elements below in solar system: a) What is solar system? b) How Did the Solar System form? 6. Investigate in the solar system the following a) Inner b) outer planets 7. Explain the following in solar system a. comets, b. Meteoroid, and c. Asteroids 8. Work in group in groups to discuss Kepler‘s laws of planetary motion and report. 9. Discuss stars pattern and constellations. 10. a) What do you mean by i. Celestial coordinates? ii. Celestial horizontal coordinate system b) Explain the following terms in celestial horizontal coordinates system i. Hour angle ii. Zenith angle 11. A) Explain what equatorial celestial coordinates system means? B) Explain the following in equatorial celestial coordinates system a. Right ascension b. Declination  41 DIFFERENT EXERCISES 1. A ray of light incident at an angle i on a prism of angle, A, passes through it symmetrically. Write an expression for the deviation, d, of the ray in terms of i and A. Hence find the value of d, if the angle of the prism is 60o and the refractive index of the glass is 1.48. 2. A beam of monochromatic light in incident normally on the refracting surface of a 60o glass prism of refractive index 1.62. Calculate the deviation caused by the prism. 3. a) Define the critical angle of a medium. b) One side of a triangular glass prism put in a pool of water of refractive index 4/3 and the other side was left open to air. A ray of light from water was incident on the prism at an angle i =21.7o. The light just grazes as it emerges out of the prism. Given that the refractive index of glass 1.52, determine the refracting angle A of the prism. 4. A monochromatic light is incident at an angle of 45o on a glass prism of refracting angle 70o in air. The emergent ray grazes the boundary of the other refracting surface of the prism. Find the refractive index of the material of glass. 5. A prism of diamond has a refracting angle of 60o. A ray of yellow light is incident at an angle of 60o on one face. Find the angle of emergence if the refractive index of diamond for yellow light is 2.42. 6. A ray of light just undergoes total internal reflection at the second face of a prism of refracting angle 60o and refractive index 1.5. What is its angle of incidence on the first face? 7. A sharp image is located 78.0mm behind a 65.0mm-focal-length converging lens. Find the object distance (a) using a ray diagram, (b) by calculation. 8. What is (a) the position, and (b) the size of the image of a 7.6cm high flower placed 1.00m from a 50.0mm focal length camera lens? 9. An object is placed 10cm from a lens of 15m of focal length. Determine the image position. 10. Two converging lenses A and B, with focal lengths fA=20cm and fB = - 25cm, are placed 80cm apart, as shown in the figure (1). An object is placed 60cm in front of the first lens as shown in figure (2). Determine (a) the position, and (b) the magnification, of the final image formed by the combination of the two lenses.  42 (a) ( b) Fig1.35: Formation of image by two lens system 11. Where a small insect must be placed if a 25cm focal length diverging lens is to form a virtual image 20cm in front of the lens? 12. Where must a luminous object be placed so that a converging lens of focal length 20cm produces an image of size four times bigger than the object (Consider the case of a real image and the case of a virtual) 13. From a real object AB we want to obtain an inverted image four times bigger than the object. We place a screen 5m away the object. Specify the kind, the position and the focus of the lens to use. Give the graphical and the algebraic. 14. In cinematography the film is located at 30m from the screen and the image has a magnification of 100. Determine the focal length of the lens used in projection. 15. An object AB of 1cm is placed at 8cm from a converging lens of focal length 12cm. Find its image (Position, nature and the size). 16. An object of 2cm is placed at 50cm from a diverging lens of focal length 10cm. Determine its image.  43 17. An object located 32.0 cm in front of a lens forms an image on a screen 8.00 cm behind the lens. (a) Find the focal length of the lens. (b) Determine the magnification. (c) Is the lens converging or diverging? 18. A movie star catches the reporter shooting pictures of her at home. She claims the reporter was trespassing. To prove her point, she gives as evidence the film she seized. Her 1.72m height is 8.25mm high on the film and the focal length of the camera lens was 210mm. How far away from the subject was the reporter standing? 19. A lighted candle is placed 33cm in front of a converging lens of focal length f1=15cm, which in turn is 55cm in front of another converging lens of focal length f2=12cm. (a) Draw a ray diagram and estimate the location and the relative size of the final image. (b) Calculate the position and relative size of the final image. Fig1.36: Image formation by two lens systems 20. When an object is placed 60cm from a certain converging lens, it forms a real image. When the object is moved to 40cm from the lens, the image moves 10cm farther from the lens. Find the focal length of this lens. 21. A converging glass lens (n=1.52) has a focal length of 40.0cm in air. Find its focal length when it is immersed in water, which has an index of refraction of 1.33. 22. Verify that the focal length f of a symmetrical biconvex lens which the two faces have a radius of curvature R and refractive index 1.5 is fmeter=Rmeter. 23. We put in contact a converging lens of focal length 20cm and a diverging lens of focal length 50cm. What are the nature, the power and the focal length of the constituted lens? 24. To a converging lens of focal length 20cm we put in contact a second lens so that the constituted system has the power of 4 diopters. Determine the nature of the second lens and calculate the focal length. 25. In a physics lab students want to determine the focal length x of a thin diverging lens. They stick to it a converging lens of 5 diopters and they use  44 the system to have a real and inverted image A‘B‘ of size equal to the one of the object AB. The distance from the object AB to the screen where they watch the image is 4m. Calculate x. 26. A thin glass lens n = 1.5 has a focal length +10cm in air. Compute its focal length in water n = 1.33. 27. A prism which has a refracting angle equals 60o and refractive index 1.5 receives a ray at an angle of incidence 45o; calculate the angle of emergence and the deviation of the ray. 28. Calculate for the same prism (Question 1) the value of minimum deviation as well as the value of i = i‘. 29. Let consider a prism made in glass of refracting angle A=59o and the refractive index 1.52. a) Calculate the deviation that makes an emerging ray with the extension of the incident ray for an incidence equal to 35o. b) Calculate the angle of minimum deviation and specify the value of the angle of the corresponding angle of incidence and refraction inside the prism. 30. Given that a prism of refracting angle A = 60o and refractive index n=√3. Let consider a ray of light falling on a prism through an angle i=90o. If it goes out the prism through an angle i‘, calculate i‘. 31. Through what angle i must fall on the prism a ray to go out through an emergence i =90o. 32. Find the refractive index of a prism A = 60o producing a minimum deviation equal to 40o. 33. A triangular glass prism with apex angle 60.0° has an index of refraction of 1.50. (a) Show that if its angle of incidence on the first surface is θ1 =48.6°, light will pass symmetrically through the prism, (b) Find the angle of deviation Dmin for θ1 = 48.6°. (c) What If? Find the angle of deviation if the angle of incidence on the first surface is 45.6°. (d) Find the angle of deviation if θ1 = 51.6°. 34..A triangular glass prism with apex angle Ф = 60.0° has an index of refraction n= 1.50. What is the smallest angle of incidence θ1 for which a light ray can emerge from the other side?  45 Fig1.37: smallest angle of incidence for ray to emerge in prism 35. A triangular glass prism with apex angle Ф has index of refraction n. What is the smallest angle of incidence θ1 for which a light ray can emerge from the other side? 2.9. End units Questions 1. A slide projector has a converging lens of focal length 20.0cm and is used to magnify the area of a slide, 5cm2 to an area of 0.8m2 on a screen. Calculate the distance of the slide from the projector lens. 2. A color slide has a picture area 2.4 cm x 3.6 cm. Find the focal length of the projection lens which will be needed to throw an image 1.2m x 1.8m on a screen 5m from the lens. 3. A projector projects an image of area 1 m2 onto a screen placed 5m from the lens. If the area of the slide is 4 m2, calculate; (i) the focal length of the projection lens. (ii) The distance of the slide from the lens 4. The focal lengths of the objective and the eye-piece of a compound microscope are 2.0 cm and 3.0 cm respectively. The distance between the objective and the eye-piece is 15.0 cm. The final image formed by the eye-piece is at infinity. The two lenses are thin. The distances in cm of the object and the image produced by the objective measured from the objective lens are respectively (a) 2.4 and 12.0 (b) 2.4 and 15.0 (c) 2.3 and 12.0 (d) 2.3 and 3.0 The answer is (a) 2.4 and 12.0cm 5. The focal lengths of the objective and eye-lens of a microscope are 1 cm and 5 cm respectively. If the magnifying power for the relaxed eye is 45, then the length of the tube is (a) 30 cm (b) 25 cm (c) 15 cm (d) 12 cm, Answer is (c) 15 cm 6. If the focal lengths of objective and eye lens of a microscope are 1.2 cm and 3 cm respectively and the object is put 1.25 cm away from the objective lens  46 and the final image is formed at infinity, then magnifying power of the microscope is (a) 150 (b) -200 (c) 250 (d) 400 The answer is (b)- 200 7. If an object subtend angle of 2° at eye when seen through telescope having objective and eyepiece of focal length fo = 60 cm and fe = 5 cm respectively than angle subtend by image at eye piece will be (a) 16° (b) 50° (c) 24° (d) 10° , Answer is (c) 24° 8. The focal lengths of the lenses of an astronomical telescope are 50 cm and 5 cm. he length of the telescope when the image is formed at the least distance of distinct vision is (a) 45 cm (b) 55 cm (c) 𝑚 (d) ) 𝑚, the answer is (d) 9. The diameter of moon is 3.5 × 103𝑘𝑚and its distance from the earth is 3.8 × 105𝑘𝑚 If it is seen through a telescope whose focal length for objective and eye lens are 4 m and 10 cm respectively, then the angle subtended by the moon on the eye will be approximately (a) 15° (b) 20° (c) 30° (d) 35°, Answer is (b) 20° 10. A telescope has an objective lens of 10 cm diameter and is situated at a distance one kilometre from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 Å, is of the order of (a) 0.5 m (b) 5 m (c) 5 mm (d) 5cm, Answer is (b) 5 m 11. A compound microscope has a magnifying power 30. The focal length of its eye-piece is 5 cm. assuming the final image to be at the least distance of distinct vision. The magnification produced by the objective will be (a) +5 (b) – 5 (c) +6 (d) – 6, Answer is (b) – 5 12. A man is looking at a small object placed at his least distance of distinct vision. Without changing his position and that of the object he puts a simple microscope of magnifying power 10 X and just sees the clear image again. The angular magnification obtained is (a) 2.5 (b) 10.0 (c) 5.0 (d) 1.0, Answer is (d) 1.0 13. The focal lengths of the objective and the eye-piece of a compound microscope are 2.0 cm and 3.0 cm respectively. The distance between the objective and the eye-piece is 15.0 cm. The final image formed by the eye- piece is at infinity. The two lenses are thin. The distances in cm of the object and the image produced by the objective measured from the objective lens are respectively (a) 2.4 and 12.0 (b) 2.4 and 15.0 (c) 2.3 and 12.0 (d) 2.3 and 3.0, The answer is (a) 2.4 and 12.0 14. (a) With the aid of a ray diagram, describe how a convex lens is used as a magnifying glass. (b) Explain why an image formed in a magnifying glass is almost free from chromatic aberration.  47 15. (a) When is a compound microscope said to be in normal use? (b) Derive an expression for the magnifying power of a compound microscope in normal use. (c) Explain why the lenses that make up a compound microscope are of short focal lengths. 16. (a) When is a telescope said to be in normal adjustment. (b) What is meant by the eye ring as applied to optical instruments. (c) What are the differences between microscope and telescopes? 17. (a) Explain why prisms are preferred to mirrors in prism binoculars. (b) State the advantages of reflecting telescopes over refracting telescopes. (c) The objective of an astronomical telescope in a normal adjustment has a diameter of 15cm and a focal length of 400cm. The eye piece has a focal length of 2.5cm. Find the magnifying power of the telescope. 18. (a) A distant objective subtending an angle of 3x 10-5 and in viewed with a reflecting telescope whose objective is a concave mirror of focal length 10m. The reflected light falls on a concave mirror placed 9.5cm from the pole of the objective which reflects the length back and a real image is formed a the pole of the objective where there is a hole. The image is viewed with a convex lens of focal length 5cm used as a magnifying glass which produces the final image at infinity. 19. How far must a 50mm focal-length camera lens be moved from its infinity setting to sharply focus an object 3m away? 20. Sue is far-sighted with a near point of 100cm. Reading glasses must have what lens power so that she can read a newspaper at a distance of 25cm? Assume the lens is very close to the eye. 21. A near-sighted eye has near and far point of 12cm and 17cm, respectively. (a) What lens power is needed for this person to see distant objects clearly, and (b) What then will be the near point? Assume that the lens is 2cm from the eye (typical for eye glasses). 22. What power contact lens is needed for an eye to see distant objects if its point is 25cm? 23. An 8cm focal-length converging lens is used as a ―jeweler‘s loupe‖, which is a magnifying glass. Estimate (a) the magnification when the eye is relaxed, and (b) the magnification if the eye is focused at its near point N=25cm. 24. A compound microscope consists of a 10X eyepiece and 50X objective 17cm apart. Determine (a) the overall magnification, (b) the focal length of each lens, and (c) the position of the object when the final image is in focus with eye relaxed. Assume a normal eye, so N = 25cm.  48 25. A near-sighted person cannot see objects clearly beyond 25.0cm (her far point). If she has no astigmatism and contact lenses are prescribed for her, what power and type of lens are required to correct her vision? 26. Microscope uses an eyepiece with a focal length of 1.4cm. Using a normal eye with a final image at infinity the tube length is 17.5cm and the focal length of the objective lens is 0.65cm. What is the magnification of the microscope? 27. The magnifying power of an astronomical telescope is 8 and the distance between the two lenses is 54cm. The focal length of eye lens and objective lens will be respectively (a) 6 cm and 48 cm (b) 48 cm and 6 cm (c) 8 cm and 64 cm (d) 64 cm and 8 cm 37 Answer is (a) 6 cm and 48 cm 3.12. End unit question 1. A machine weighing 1500 N is supported by two chains attached to some point on the machine. One of these ropes goes to a nail in the wall and is inclined at 30° to the horizontal and other goes to the hook in ceiling and is inclined at 45° to the horizontal. Find the tensions in the two chains. Fig 3.32: Showing forces acting on a body in different directions 2. A uniform horizontal beam 5.00 m long and weighing 3 × 102𝑁 is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 530 with the horizontal. If a person weighing 6 × 102𝑁 stands 1.50 m from the wall, find the magnitude of the tension in the cable and the force 𝑅 exerted by the wall on the beam.  49 Figure 3.33: Showing the forces developed in a bar when subjected to tension 3. A box of weight W rests on the sloping plank. The coefficient of static friction between the surfaces is 0.25. If tIn static, a body is said to be in equilibrium when the force system acting upon it has a zero resultant. Figure 3.34: Showing forces acting on a body in different directions 4. To determine the values of P and F, when the body M is in equilibrium, use the projection of the force on the x and y-axis. The slope of the plank is gradually increased, at what angle of the slope will the box begin to slide? 5. Determine analytically the magnitude and direction of the resultant of the following four forces acting at a point. (i) 10 kN pull N 30° E; (ii) 20 kN push S 45° W; (ii) 20 kN push S 45° W; (iii) 5 kN push N 60° W; (iv) 15 kN push S 60° E. (iv) 15 kN push S 60° E.  50 Figure 3.35: Showing forces acting on a body in different directions 6. A seesaw consisting of a uniform board of mass M =10kg and length l=2m supports a father and daughter with masses mf and md, 50 and 20kg respectively as shown in the Figure. The support (called the fulcrum) is under the centre of gravity of the board. The father is a distance d from the centre, and the daughter is at distance l/2 from the centre. (a) Determine the magnitude of the upward force (reaction) n exerted by the support on the board. (b) Determine where the father should sit to balance the system. Figure 3.36: A seasaw 7. A person holds a 50.0N sphere in his hand. The forearm is horizontal, as shown in Figure. The biceps muscle is attached 3.0cm from the joint, and the sphere is  51 35.0cm from the joint. Find the upward force exerted by the biceps on the forearm and the downward force exerted by the upper arm on the forearm and acting at the joint. Neglect the weight of the forearm. Figure 3.37: Showing forces on the arm lifting a heavy mass 8. A uniform horizontal beam with a length of 8.00m and a weight of 200N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of 53.0° with the beam. If a 600N person stands 2.00m from the wall, find the tension in the cable as well as the magnitude and direction of the force exerted by the wall on the beam. 9. Calculate the magnitudes FA and FB of the tensions in the two cords that are connected to the vertical cord supporting the 200kg chandelier in the figure. Figure 3.38: Indicating forces developed when a mass is held by two strings 10. A uniform 1500kg beam, 20m long, supports a 15,000kg printing press 5 from the right support column, see the figure. Calculate the force on each of the vertical support columns.  52 Figure 3.39: Indicating forces developed in a beam 11. The bar in the figure is being used as a lever to pry up a large rock. The small rock acts as a fulcrum (pivot point). The force FP required at the long end of the bar can be quite a bit smaller than the rock‘s weight mg, since it is torques that balance in the rotation about the fulcrum. If, however, the leverage isn‘t sufficient, and the large rock isn‘t budged, what are the two ways to increase the leverage? SOLVED EXERCISES Q1. Explain the variation of acceleration of gravity with: 2 marks a) height b) depth c) latitude d) earth rotation Q2. List and explain four types of satellites orbits based on the angle of inclination. 4 marks Q3. What is the difference between fluorescence and phosphorescence? 4 marks Q4. Compare bandwidth and frequency. 3 marks  53 Q5. Match each sentence with the corresponding correct letter from the figure below. 3 marks 1.) The emission line with the longest wavelength. 2.) The absorption line with the shortest wavelength. 3.) The emission line with the highest energy. 4.) The absorption line with the lowest energy. 5.) The emission of line with the lowest frequency. 6.) The line corresponding to the ionization energy of hydrogen. Q6. a.) What are the three notational methods for describing the behavior of gates and circuits? 3 marks b.) Show the behavior of the following circuit with a truth table: 4 marks A B C Q7. What are the principles of Cellular Architecture? 2 marks Q8. After describing the interaction of Electromagnetic radiations with atmosphere, explain blue color of the sky and what makes the red sunset and sunrise. 5 marks Q9. Three point charge, q1=1µC, q2=-2µC, and q3=3µC are fixed at the positions shown in Figure. (a) What is the potential at point P at the corner of the rectangle? 3 marks (b) What is the total potential energy of q1, q2, and q3? 3 marks (c) How much work would be needed to bring a charge q4=2.5µC from infinity and to place it at P? 3 marks  54 Q10. Answer the following questions using the HR diagram below. 11 marks 1. What factor affects the color of a star? 2. What factors affect the luminosity of a star? 3. What is the approximate surface temperature of the sun? 4. Is the temperature of white dwarf stars higher or lower than red super giant? 5. What is the color of the stars with the highest surface temperature? 6. What is the color of the stars with the lowest surface temperature? 7. List the color of the stars from hottest to coldest 8. Most of the stars on the HR diagram are classified as which type of star? 9. What type of star has high temperature but low luminosity? 10. What type of star has high temperature and high luminosity? 11. What type of star has a low temperature but a high luminosity? 12. What type of star has a low temperature and a low luminosity? 13. Plot the star A-E. Once plotted determine their color and type. letter temperature luminosity color Types of star A 6000 k 10-1 B 20,000 k 106 C 20,000 k 10-2 D 2,500 k 106  55 E 4,000 k 102 ANSWERS: Q1. Explain the variation of acceleration of gravity with: 2 marks e) height This shows that acceleration due to gravity decreases with increase in altitude. f) depth The acceleration due to gravity decreases with increase in depth. g) Latitude Hence, it is inversely proportional to the square of the radius. It is least at the equator and maximum at the poles, since the equatorial radius (6378.2 km) is more than the polar radius (6356.8 km) h) earth rotation At poles; φ = 90° g‘ = g. So, at equator the effect is highest so the gravity is lowest and at the poles gravity remains unaffected of rotation. Q2. List and explain four types of satellites orbits based on the angle of inclination. 4 marks  56 There are four types of orbits based on the angle of inclination. Equatorial orbit – Angle of inclination is either zero degrees or 180 degrees. Polar orbit - Angle of inclination is 90 degrees. Prograde orbit - Angle of inclination lies between zero and 90 degrees. Retrograde orbit - Angle of inclination lies between 90 and 180 degrees. Q3. What is the difference between fluorescence and phosphorescence? 4 marks Q4. Compare bandwidth and frequency. 3 marks COMPARISON BETWEEN BANDWIDTH AND FREQUENCY: Bandwidth Frequency 1. The rate at which The number of complete Definition data is transferred cycles per second in from one network to alternating current direction  57 another 2. The difference between the highest frequency signal component and the lowest-frequency signal component Time measurement Per second Per second Science and engineering to specify the rate of oscillatory and vibratory phenomena, Computing, signal Used in such as mechanical processing vibrations, audio (sound) signals, radio waves, and light Standard Unit Hertz; Hz Hertz; Hz Kilohertz, megahertz, Kilohertz, megahertz, Larger Units gigahertz, terahertz gigahertz, terahertz Q5. Match each sentence with the corresponding correct letter from the figure below. 3 marks 7.) The emission line with the longest wavelength. E 8.) The absorption line with the shortest wavelength. A 9.) The emission line with the highest energy. D 10.) The absorption line with the lowest energy. B 11.) The emission of line with the lowest frequency. E 12.) The line corresponding to the ionization energy of hydrogen. A Q6. a.) What are the three notational methods for describing the behavior of gates and circuits? Boolean expressions, logic diagrams, and truth tables. 3marks b.) Show the behavior of the following circuit with a truth table: 4 marks A B C  58 A B C AB (BC)’ C’ (AB+C’)’ (BC)’ + (AB+C’)’ 0 0 0 0 1 1 0 1 0 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 0 0 1 1 1 0 0 0 1 1 0 1 1 0 1 0 1 0 1 1 1 1 0 1 1 1 0 1 1 1 1 1 0 0 0 0 Q7. What are the principles of Cellular Architecture? 2 marks  Low power Transmitters and Coverage Zones.  Frequency Reuse.  Cell splitting to increase Capacity.  Hand off and Central Control. Q8. After describing the interaction of Electromagnetic radiations with atmosphere, explain blue color of the sky and what makes the red sunset and sunrise. 5 marks Before radiation used for remote sensing reaches the Earth's surface it has to travel through some distance of the Earth's atmosphere. Particles and gases in the atmosphere can affect the incoming light and radiation. These effects are caused by the mechanisms of transmission, scattering and absorption. Scattering occurs when particles or large gas molecules present in the atmosphere interact with and cause the electromagnetic radiation to be redirected from its original path. How much scattering takes place depends on several factors including the wavelength of the radiation, the abundance of particles or gases, and the distance the radiation travels through the atmosphere. There are three (3) types of scattering which take place.  Rayleigh scattering  Mie scattering  Nonselective scattering Rayleigh scattering occurs when particles are very small compared to the wavelength of the radiation. These could be particles such as small specks of dust or nitrogen and oxygen molecules. Rayleigh scattering causes shorter wavelengths of energy to be scattered much more than longer wavelengths. Rayleigh scattering is the dominant scattering mechanism in the upper atmosphere. The fact that the sky appears "blue" during the day is because of this phenomenon. As sunlight passes through the atmosphere, the shorter wavelengths (i.e. blue) of the visible spectrum are scattered  59 more than the other (longer) visible wavelengths. At sunrise and sunset the light has to travel farther through the atmosphere than at midday and the scattering of the shorter wavelengths is more complete; this leaves a greater proportion of the longer wavelengths to penetrate the atmosphere. Mie scattering occurs when the particles are just about the same size as the wavelength of the radiation. Dust, pollen, smoke and water vapour are common causes of Mie scattering which tends to affect longer wavelengths than those affected by Rayleigh scattering. Mie scattering occurs mostly in the lower portions of the atmosphere where larger particles are more abundant, and dominates when cloud conditions are overcast. The final scattering mechanism of importance is called nonselective scattering. This occurs when the particles are much larger than the wavelength of the radiation. Water droplets and large dust particles can cause this type of scattering. Nonselective scattering gets its name from the fact that all wavelengths are scattered about equally. This type of scattering causes fog and clouds to appear white to our eyes because blue, green, and red light are all scattered in approximately equal quantities (blue+green+red light = white light). Absorption is the other main mechanism at work when electromagnetic radiation interacts with the atmosphere. In contrast to scattering, this phenomenon causes molecules in the atmosphere to absorb energy at various wavelengths. Ozone, carbon dioxide, and water vapour are the three main atmospheric constituents which absorb radiation.  60 Ozone serves to absorb the harmful (to most living things) ultraviolet radiation from the sun. Without this protective layer in the atmosphere our skin would burn when exposed to sunlight. You may have heard carbon dioxide referred to as a greenhouse gas. This is because it tends to absorb radiation strongly in the far infrared portion of the spectrum - that area associated with thermal heating - which serves to trap this heat inside the atmosphere. Water vapour in the atmosphere absorbs much of the incoming longwave infrared and shortwave microwave radiation (between 22µm and 1m). The presence of water vapour in the lower atmosphere varies greatly from location to location and at different times of the year. For example, the air mass above a desert would have very little water vapour to absorb energy, while the tropics would have high concentrations of water vapour (i.e. high humidity). Because these gases absorb electromagnetic energy in very specific regions of the spectrum, they influence where (in the spectrum) we can "look" for remote sensing purposes. Those areas of the spectrum which are not severely influenced by atmospheric absorption and thus, are useful to remote sensors are called atmospheric windows. By comparing the characteristics of the two most common energy/radiation sources (the sun and the earth) with the atmospheric windows available to us, we can define those wavelengths that we can use most effectively for remote sensing. The visible portion of the spectrum, to which our eyes are most sensitive, corresponds to both an atmospheric window and the peak energy level of the sun. Explain blue color of the sky and what makes the red sunset and sunrise. Light from the Sun contains all colours of the rainbow. When we look directly at the Sun, we see all of the light from it, so it appears white