M = f objective f ocular. (see page 374 of the textbook). The minus sign in this formula indicates that the image is upside-down.

Transcription

1 PHY 309 L. Solutions for homework set # 9. Textbook problem E.16 at the end of chapter 17: The angular magnification of a telescope depends on the ratio of focal distances of the objective lens (or mirror) and the ocular (eyepiece) lens: M = f objective f ocular (1) (see page 374 of the textbook). The minus sign in this formula indicates that the image is upside-down. The telescope in question (probably intended for an amateur astronomer) has f objective = 40 cm and f ocular = 2.5 cm, so its angular magnification is 40 cm M = 2.5 cm = 16. (2) Non-textbook problem #I: (a) When you look at some object through a negative lens, you see a virtual image which is closer to you than the actual object, d img < d obj. For nearby objects, the difference is small, but becomes larger for more distat object, and even for the far-away objects the images are never further away than the focal length f of the lens. This effect would help a near-sighted person whose eyes have hard time focusing on distant objects but would be able to focus on the closer-by virtual images. (b) Positive lenses have precisely the opposite effect: When you look at an object through a positive lens, you see a virtual image which is more distant than the object, d img > d obj. Consequently, a positive lens would help a far-sighted person whose eyes have hard time focusing on nearby objects but can focus on the more distant virtual images. 1

2 (c) When people age, the longest distance to which their eyes can focus remains the same, but the shortest distance becomes longer. Thus many people have normal vision while they are young but become far-sighted when they reach their forties of fifties. But the people who are slightly near-sighted while they are young can become both near-sighted and far-sighted when they age. That is, their eyes can focus on objects at intermediate distances, say from 1.5 feet to 15 feet, but cannot focus on object which are either too close or too far. To read a book that s only 1 foot away from their eyes they need positive lenses. But for driving, to see the distant object well they need negative glasses. (d) NO. People who need negative lenses for reading are strongly nearsighted. The light rays entering their eyes cross way before reaching the retina even for nearby objects like a book 10 inches away from their faces. For the more distant objects the effect is even stronger, so they need more negative lenses to correct the problem. For example, a person who needs 2 lenses to read would need 6 lenses to walk or to drive (assuming s/he could get a license). On the other hand, people who need positive lenses for driving or walking which require looking at distant objects are strongly farsighted. The light rays entering their eyes do not cross by the time they reach the retina, even if they come from far-away objects. For closer objects (like reading materials) the effect would be stronger, so such people would need more positive lenses to correct the problem. For example, a person who need +2 glasses to drive would need +5 glasses to read. There is no way anybody can be simultaneously too nearsighted to read and too farsighted to walk. Non-textbook problem #II: (a) The object is just a little more distant from the objective lens than its focal distance, so it makes a real image on the other side of the lens. That is, above the lens, since the specimen is usually held below the objective. The distance from the lens to the image follows from 1 d o 1 d i = 1 f (3) 2

3 (cf. eq. (4) of my notes), thus d i = d o f d o f = 13.0 mm 12.0 mm 13.0 mm 12.0 mm = 156 mm. (4) Altogether, the real image in the objective lens is located 156 mm above the objective lens. (b) The magnification of a real image in the objective lens is m objective h i h o = d i d o = f d o f = 12.0 mm 13.0 mm 12.0 mm = 12. (5) The minus sign here indicates that the real image is up-side down. (c) The ocular (the eyepiece) makes a virtual image of the real image made by the objective. This works as if the image made by the objective was an object on its own right. That is, the objective s image and its virtual image in the ocular are on the same side of the lens below it, since we look from the above and their distances d o and d i from the ocular are related as 1 d o 1 d i = 1 f. (6) Thus, given the focal length f = 25.0 mm of the ocular and the desired distance d i = 25 cm = 250 mm of the virtual image from the lens, the distance d o of the objective s image from the ocular lens should be / ( d 1 o = ) f d i = d i f d i + f = 250 mm 25.0 mm 250 mm mm = 22.7 mm. (7) In other words, to put the final virtual image 25 cm from the observer s eye, the ocular lens should be put 22.7 mm above the image made by the objective lens. In light of part (a), this means that the ocular lens should be about 189 mm above the objective lens. (d) The magnification of a virtual image in the ocular lens is m ocular h i h o = + d i d o 250 mm = mm = +11. (8) The plus sign here indicates that the virtual image has the same orientation as its object (which happens to be the image in the objective lens). 3

4 (e) The net magnification of the microscope is m net = m objective m ocular = ( 12) (+11) = 132. (9) That is the final image is upside-down and 132 times bigger than the original object the infusoria. This, if the infusoria is 0.2 mm long, its magnified image has length L = mm = 26.5 mm. (10) Or in round numbers, the image is about one inch long. Non-textbook problem III: In a chemical reaction the net mass is conserved the net mass of all the reaction products is equal to the net mass consumed by the reaction. However the net mass comprises all inputs or all outputs to the reaction, including the gases. Thus, when a piece of wood is burned, the input includes both the wood and the oxygen consumed by the burning, while the output includes not just the ash but also the gaseous product of combustion, mostly carbon dioxide and water vapor. Consequently, the mass conservation means M wood + M oxygen = M ash + M CO2 + M H2O + M other gases. (11) Together, CO 2, water vapor, and other combustion gases weight more than then the oxygen consumed by the burning, so the ash weighs less than the wood. Non-textbook problem IV: Even for a pure element like oxygen or hydrogen, the molecules are quite different from atoms. In a chemical reaction, the atoms remain the same but the molecules change into different molecules. For example, the reaction 2 H 2 + O 2 2 H 2 O (12) does not change the net number of oxygen atoms or the net number of hydrogen atoms. At the end of the reaction, we still have all the oxygen atoms or hydrogen atoms we had in the beginning, they just belong to different molecules. 4

5 However, the atoms reorganize themselves into different molecules. We start with hydrogen molecules H 2 each comprising 2 hydrogen atoms and oxygen molecules O 2 each comprising 2 oxygen atoms. In the course of the reaction, all these molecules are broken and the atoms recombine into water molecules H 2 O, each comprising 2 hydrogen atoms and one oxygen atom. Thus, the number of hydrogen molecules decreases (eventually to zero) but the number of hydrogen atoms stays the same. Likewise, the number of oxygen molecules decrease, but the number of oxygen atoms stays the same. Textbook problem E.2 at the end of chapter 18: Given the atomic masses of carbon and oxygen, we can say that the mass of one CO 2 molecule comprises 12 atomic mass units (amu) of carbon and 2 16 = 32 atomic mass units or oxygen. Consequently, carbon and oxygen combine into carbon dioxide in the mass proportion 12 : 32. For example, 12 grams of carbon to 32 grams of oxygen. In particular, 64 grams of oxygen would combine with of carbon. Textbook problem E.4 at the end of chapter 18: g = 24 grams (13) 32 One molecule of alumina (aluminum oxide) Al 2 O 3 contains 2 atoms of aluminum and 3 atoms of oxygen. The atomic weight of aluminum is 27, which means the mass of one aluminum atom is 27 atomic mass units (amu). Likewise, the mass of an oxygen atom is 16 atomic mass units. Hence, the net mass of one alumina molecule is m(al 2 O 3 ) = 2 m(al atom) + 3 m(o atom) = 2 27 amu amu = 102 amu. Out of these 102 amu, 2 27 = 54 amu are made of aluminum and 3 16 = 48 amu are made of oxygen. The same ratio will apply to any amount of aluminum oxide: (14) In Al 2 O 3, m(aluminum) m(oxygen) = 2 µ(al) 3 µ(o) = = (15) Therefore, to oxidize 54 arbitrary mass units of aluminum and make AL 2 O 3 one needs 48 5

6 similar units of oxygen. This will work for any mass units grams, pounds, tons, whatever as long as the aluminum and the oxygen masses are measured in the same units. In particular, oxidizing 54 grams of aluminum takes 48 grams of oxygen. 6