Practice final exam, fully corrected (PHYS 101)

a) One conversion, used twice. $270$ km/h $= 270/3.6 = 75.0$ m/s and $150$ km/h $= 150/3.6 = 41.67$ m/s. The equation that avoids the time is $v_{f}^{2} = v_{i}^{2} + 2a\,\Delta x$, and with $v_{i} = 0$ it gives $a = \frac{v_{f}^{2}}{2\Delta x} = \frac{75.0^{2}}{2(1800)} = \frac{5625}{3600} = 1.5625$ m/s$^2$. Then $t = v_{f}/a = 75.0/1.5625 = 48.0$ s. Check it a second way with the average speed: the acceleration is constant and the start is from rest, so the average speed is $75.0/2 = 37.5$ m/s and $1800/37.5 = 48.0$ s. The trap here is not the physics, it is converting once and then writing $270$ again in the next line: the number $270$ has no place in any equation of this course.

b) Two phases, each with its own constant acceleration, so two separate applications of the same equation. Accelerating up to $41.67$ m/s uses $\Delta x_{1} = \frac{41.67^{2}}{2(1.5625)} = 555.6$ m. Braking from $41.67$ m/s to rest at $3.5$ m/s$^2$ uses $\Delta x_{2} = \frac{41.67^{2}}{2(3.5)} = 248.0$ m. Total $803.6$ m, so $1800 - 804 = 996$ m of runway are still ahead when the aircraft stops. Notice that the braking phase is shorter than the accelerating one even though both start or end at the same speed: the brakes are more than twice as strong as the engines, and distance goes as the inverse of the acceleration at fixed speed. The solution figure draws the whole abort as a velocity time graph, where the two distances are simply the areas of the two triangles.

c) The climb is a vector at $8.0^{\circ}$ from the horizontal, so the two components come from a right triangle: $v_{y} = 85\sin 8.0^{\circ} = 85(0.13917) = 11.83$ m/s and $v_{x} = 85\cos 8.0^{\circ} = 85(0.99027) = 84.17$ m/s. The climb rate is the vertical component alone, so $t = 900/11.83 = 76.1$ s, and during that time the ground distance is $84.17(76.1) = 6.40 \times 10^{3}$ m. Check without the time at all: the path is a straight line, so ground distance $= 900/\tan 8.0^{\circ} = 900/0.14054 = 6404$ m. The two routes agree, which is the sign that the components were taken with the right trigonometric function. Writing $v_{y} = 85\cos 8.0^{\circ}$ would give a climb rate of $84$ m/s, about seven times what any airliner can do: an order of magnitude check catches that mistake before the marker does.

d) Same final speed, longer runway, so $a = \frac{v_{f}^{2}}{2\Delta x}$ gives $a = \frac{5625}{2(2100)} = 1.339$ m/s$^2$ and $t = 75.0/1.339 = 56.0$ s. Read the two formulas as proportions rather than recomputing: at fixed $v_{f}$, $a$ is inversely proportional to $\Delta x$, so $a' = 1.5625 \times \frac{1800}{2100} = 1.339$ m/s$^2$; and $t = 2\Delta x / v_{f}$ is directly proportional to $\Delta x$, so $t' = 48.0 \times \frac{2100}{1800} = 56.0$ s. Seventeen per cent more runway buys seventeen per cent more time and costs fourteen per cent of the acceleration.

a) $v_{x} = 16.0\cos 20.0^{\circ} = 15.04$ m/s and $v_{y0} = 16.0\sin 20.0^{\circ} = 5.472$ m/s. The horizontal motion has no acceleration, so the net is reached when $x = v_{x}t = 9.00$, that is $t = 9.00/15.04 = 0.599$ s. The horizontal column is always the one that carries the clock in this chapter, because it is the only one with a constant speed.

b) With the same clock, the vertical column gives $y = 2.10 + 5.472(0.599) - 4.90(0.599)^{2} = 2.10 + 3.276 - 1.756 = 3.620$ m. The clearance is $3.620 - 2.43 = 1.19$ m. Note what the question did NOT ask: the ball is nowhere near the top of its flight at that instant, and computing the apex height would answer a different question. The apex is reached at $t = v_{y0}/g = 0.558$ s, slightly before the net, at a height of $3.628$ m, so the ball crosses the net a few centimetres past its highest point and is already coming down.

c) Landing means $y = 0$, so $4.90t^{2} - 5.472t - 2.10 = 0$. The discriminant is $5.472^{2} + 4(4.90)(2.10) = 29.95 + 41.16 = 71.11$, its square root is $8.432$, and the physical root is the positive one, $t = \frac{5.472 + 8.432}{9.80} = 1.419$ s. The negative root, $-0.302$ s, is the instant at which a ball launched from the floor would have passed through the serve point, which never happened. Then $x = 15.04(1.419) = 21.3$ m, and the end line is at $18.0$ m: the serve is out by $3.3$ m. Clearing the net and landing in are two different questions, and this serve answers yes to one and no to the other.

d) At landing $v_{y} = 5.472 - 9.80(1.419) = -8.432$ m/s, the minus sign saying downwards, while $v_{x}$ is still $15.04$ m/s because nothing ever acted horizontally. So $v = \sqrt{15.04^{2} + 8.432^{2}} = 17.24$ m/s and the angle below the horizontal is $\arctan(8.432/15.04) = 29.3^{\circ}$. The second route needs no time at all: between the serve point and the floor the speed obeys $v^{2} = v_{0}^{2} + 2g h = 16.0^{2} + 2(9.80)(2.10) = 256 + 41.16 = 297.2$, so $v = 17.24$ m/s. The two agree, and the second route is the energy idea of Part B arriving a chapter early.

a) The crate is not pushed by anything: the only force that can slow it down is the friction of the bed, so the crate can follow the truck only as long as the required force stays within what static friction can supply. $ma \le \mu_{s}mg$ gives $a \le \mu_{s}g = 0.55(9.80) = 5.39$ m/s$^2$, and the mass cancels, so a heavier crate is no safer. From $90$ km/h $= 25.0$ m/s the shortest such stop is $\Delta x = \frac{25.0^{2}}{2(5.39)} = 58.0$ m. Anything shorter and the load comes forward.

b) At $7.0$ m/s$^2$ the bed cannot hold the crate, so the crate slides and the friction switches to kinetic: the crate decelerates at only $\mu_{k}g = 0.42(9.80) = 4.12$ m/s$^2$. Both start at $25.0$ m/s. The truck stops after $\frac{25.0^{2}}{2(7.0)} = 44.6$ m; the crate needs $\frac{25.0^{2}}{2(4.12)} = 75.9$ m, sliding on the stationary bed for the last part of it. The crate therefore advances $75.9 - 44.6 = 31.3$ m relative to the truck, which is several times the length of any bed: it goes through the cab. The trap is to keep $\mu_{s}$ once the sliding has started, which would make the crate decelerate faster than it can.

c) At the top of the hill the centre of the circle is BELOW the car, so the net force must point downwards: $mg - N = \frac{mv^{2}}{R}$. Contact is lost when $N = 0$, that is $v = \sqrt{gR} = \sqrt{9.80(42)} = 20.3$ m/s, or $73.0$ km/h. Again the mass cancels. Compare this with a bucket swung in a vertical circle, where the same condition $N = 0$ gives a MINIMUM speed, below which the load falls away from the path. Here the road can only push, never pull, so the same equation reads as a MAXIMUM: above $73.0$ km/h the car leaves the road. Which of the two it is depends on which side of the path the support sits, not on the algebra.

d) For the passenger, $mg - N = \frac{mv^{2}}{R}$, so $N = m\left(g - \frac{v^{2}}{R}\right) = 68\left(9.80 - \frac{15.0^{2}}{42}\right) = 68(9.80 - 5.357) = 68(4.443) = 302$ N. As a fraction of the weight $mg = 666$ N, that is $4.443/9.80 = 0.453$: the seat pushes with only forty five per cent of its usual force, which is the lift in the stomach a passenger feels over a hump. Writing $N = mg$ here is the standard loss of marks: $N$ equals $mg$ only when the acceleration perpendicular to the surface is zero, and on a curved road it never is.

e) Newton's second law along the radius for Io: $\frac{GMm}{r^{2}} = \frac{4\pi^{2}mr}{T^{2}}$, the mass of Io cancels, and $M = \frac{4\pi^{2}r^{3}}{GT^{2}}$. With $T = 1.77(86400) = 1.529\times 10^{5}$ s and $r^{3} = 7.515\times 10^{25}$ m$^3$, the numerator is $2.967\times 10^{27}$ and the denominator is $6.67\times 10^{-11}(2.339\times 10^{10}) = 1.560$, so $M = 1.90\times 10^{27}$ kg, about $319$ times the mass of the Earth. The orbital speed follows from the geometry alone: $v = \frac{2\pi r}{T} = 1.73\times 10^{4}$ m/s, that is $17.3$ km/s. The lesson of this scene is that the mass of a planet is never weighed, it is read off the orbit of something that goes round it, and the orbiting body's own mass never enters the answer.

a) $E_{p} = \frac{1}{2}(900)(0.0550)^{2} = \frac{1}{2}(900)(0.003025) = 1.361$ J. The factor one half is not a convention: the force needed to hold the spring is $F = kx$, so it grows from $0$ at the start of the compression to $kx$ at the end. The work is the area under the force against displacement graph, and that area is a TRIANGLE of base $x$ and height $kx$, not the rectangle $kx\cdot x$. Using the rectangle doubles every energy in this question.

b) The rail is horizontal and frictionless, so every joule stored becomes kinetic: $\frac{1}{2}mv^{2} = 1.361$ gives $v = \sqrt{\frac{2(1.361)}{0.0800}} = \sqrt{34.03} = 5.834$ m/s. Notice what was not needed: the time of the push, the length of the plunger, the force at any instant. That is the whole point of an energy method.

c) Three terms, written once and for all. Height gained: $\Delta h = 0.45\sin 12.0^{\circ} = 0.0936$ m, so $\Delta E_{p} = mg\Delta h = 0.0800(9.80)(0.0936) = 0.0734$ J. Friction: the normal force on a slope is $N = mg\cos\theta = 0.0800(9.80)(0.97815) = 0.7669$ N, so $f = \mu_{k}N = 0.0767$ N and the energy it removes is $f d = 0.0767(0.45) = 0.0345$ J. What is left is kinetic: $1.361 - 0.0734 - 0.0345 = 1.253$ J, hence $v = \sqrt{\frac{2(1.253)}{0.0800}} = 5.598$ m/s. The audit reads $5.39$ per cent to height, $2.54$ per cent to friction, $92.08$ per cent still kinetic. The common error is to use $N = mg$ on the slope, which raises the friction loss by two per cent of itself here, and by a great deal more on a steep ramp.

d) Now $E_{p} = \frac{1}{2}(900)(0.0200)^{2} = 0.180$ J. Going up, gravity and friction both oppose the motion, and both are constant along the slope, so the energy is spent at a fixed rate per metre: $mg\sin\theta + \mu_{k}mg\cos\theta = 0.1630 + 0.0767 = 0.2397$ N per metre. The ball stops after $d = 0.180/0.2397 = 0.751$ m, well inside the $1.20$ m ramp. Does it stay? Compare the pull down the slope, $mg\sin\theta = 0.163$ N, with the largest static friction available, $\mu_{s}N = 0.15(0.7669) = 0.115$ N. The pull wins, so the ball slides back. Coming down, gravity gives and friction still takes: $(0.1630 - 0.0767)(0.751) = 0.0648$ J, so it returns at $v = \sqrt{\frac{2(0.0648)}{0.0800}} = 1.273$ m/s, against the $2.121$ m/s it left with. Only $36.0$ per cent of the stored energy survives the round trip, and friction took the rest, twice.

a) The skid happens AFTER the collision, so it is a separate problem with one force, friction. Over the $14.0$ m the friction removes all the kinetic energy: $\frac{1}{2}Mv^{2} = \mu_{k}Mgd$, the total mass cancels, and $v = \sqrt{2\mu_{k}gd} = \sqrt{2(0.70)(9.80)(14.0)} = \sqrt{192.1} = 13.86$ m/s. That a $2500$ kg wreck and a bicycle would skid the same distance from the same speed is not a paradox: a heavier body carries more energy and is held by more friction, in the same proportion.

b) The collision itself is where momentum, and only momentum, is conserved: $m_{A}v_{A} = (m_{A}+m_{B})v$, so $v_{A} = \frac{2500}{1400}(13.86) = 24.75$ m/s. In the units of the traffic code that is $24.75(3.6) = 89.1$ km/h, about $19$ km/h over the $70$ km/h limit. Work backwards through the two stages in the right order, skid first and collision second, and never mix them.

c) Before: $\frac{1}{2}(1400)(24.75)^{2} = 4.288\times 10^{5}$ J. Just after: $\frac{1}{2}(2500)(13.86)^{2} = 2.401\times 10^{5}$ J. The loss is $1.886\times 10^{5}$ J, that is $44.0$ per cent, and it went into crushing metal, heat and sound. The check is worth remembering: for a perfectly inelastic collision with the target at rest the fraction lost is exactly $\frac{m_{B}}{m_{A}+m_{B}} = \frac{1100}{2500} = 0.440$, independent of the speed. A heavy car hitting a light one loses little; a light car hitting a heavy one loses almost everything.

d) The expert writes $\frac{1}{2}(1400)v_{A}^{2} = 2.401\times 10^{5}$ and gets $v_{A}^{2} = 343.0$, so $v_{A} = 18.52$ m/s $= 66.7$ km/h, just under the limit. The number is arithmetically correct and physically worthless: the two cars lock together, which is the definition of a perfectly inelastic collision, and $44$ per cent of the kinetic energy leaves the problem during the few hundredths of a second of the impact. Momentum survives that impact because the road exerts no significant horizontal impulse over so short a time; kinetic energy does not. Conserving the wrong quantity here is worth $22$ km/h, and an acquittal.

a) Take torques about the HINGE, because that axis deletes the two unknown components of the hinge force and leaves $T$ alone. Only the component of the cable tension perpendicular to the beam turns it, and that is $T\sin 35^{\circ}$, applied at $3.0$ m. Against it: the weight of the beam, $40(9.80) = 392$ N at the centre, $1.5$ m out, and the sign, $25(9.80) = 245$ N at $2.4$ m. So $T\sin 35^{\circ}(3.0) = 392(1.5) + 245(2.4) = 588 + 588 = 1176$ N m, giving $T\sin 35^{\circ} = 392$ N and $T = 392/0.5736 = 683$ N. Using $T$ itself instead of $T\sin 35^{\circ}$, or using $\cos$, is the most expensive single slip of the chapter.

b) Now the forces, in two directions. The cable pulls the end of the beam towards the wall with a horizontal component $T\cos 35^{\circ} = 683(0.8192) = 560$ N, so the hinge must push the beam AWAY from the wall with the same $560$ N. Vertically, $H_{y} + T\sin 35^{\circ} = (40+25)(9.80) = 637$ N, and since $T\sin 35^{\circ} = 392$ N exactly, $H_{y} = 245$ N upwards. The magnitude is $\sqrt{560^{2} + 245^{2}} = 611$ N, at $\arctan(245/560) = 23.6^{\circ}$ above the horizontal. It does NOT point along the beam: a hinge is not a two force member, and only a rod with forces at its two ends only would have its force along its own line.

c) With the cable gone, nothing balances the torque of the two weights about the hinge, and that torque is still $1176$ N m at the instant of the break, because nothing has moved yet. The moment of inertia about the hinge is $I = \frac{1}{3}(40)(3.0)^{2} + 25(2.4)^{2} = 120 + 144 = 264$ kg m$^2$. Then $\alpha = \frac{\sum\tau}{I} = \frac{1176}{264} = 4.45$ rad/s$^2$. Adding the two masses and treating the whole thing as a point body at some average distance gives a different and wrong number: in rotation a mass counts by the SQUARE of its distance.

d) $a = \alpha r$, so the free end starts down at $4.45(3.0) = 13.4$ m/s$^2$ and the sign at $4.45(2.4) = 10.7$ m/s$^2$. Both are larger than $g = 9.80$ m/s$^2$. The crossover is at $r = g/\alpha = 9.80/4.45 = 2.2$ m: beyond that point the beam is driven down faster than gravity alone could do it, because the inner part of the beam is being held back by the hinge and pushes the outer part down through the rigid body. Put a coin on the beam at $2.8$ m and cut the cable: the beam drops away from underneath the coin, which falls behind at a mere $9.80$ m/s$^2$. That is the classroom demonstration this question is built on.

a) $T = 2\pi\sqrt{\frac{0.9930}{9.80}} = 2\pi\sqrt{0.101327} = 2\pi(0.318318) = 2.000$ s, so $f = 1/T = 0.500$ Hz. The clock keeps time. Two things do NOT appear in that formula and are worth saying out loud, because half the marks of this chapter live there: the mass of the bob, and the amplitude. A heavier bob and a wider swing both leave the period alone.

b) Lengthening by $0.040$ per cent means $L' = 1.00040L$. Since $T \propto \sqrt{L}$, the period grows by $\sqrt{1.00040} = 1.000200$, so $T' = 2.0004$ s. Do not stop there: the question asks for seconds lost, not for a period. In one real day of $86400$ s the clock displays $86400 \times \frac{T}{T'} = 86400(0.999800) = 86383$ s, so it loses $17.3$ s. A useful shortcut for the checking pass: a fractional change of $L$ gives half as large a fractional change of $T$, and half of $0.040$ per cent of a day is $0.00020(86400) = 17$ s.

c) In Quito $T'' = 2\pi\sqrt{\frac{0.9930}{9.771}} = 2.0030$ s, and the display falls behind by $86400\left(1 - \frac{2.0000}{2.0030}\right) = 128$ s, over two minutes a day. To fix it the rod must be shortened until the period is $2.000$ s again with the smaller $g$: $L_{new} = \frac{gT^{2}}{4\pi^{2}} = \frac{9.771(4.000)}{39.478} = 0.99006$ m, that is $2.9$ mm shorter. The half rule checks it: $g$ dropped by $0.296$ per cent, so $L$ must drop by the same $0.296$ per cent, which is $0.9930(0.00296) = 2.9$ mm. The trap is to change the city and keep the old $g$ in the formula, which is the single most common way to lose this question.

d) First route, the oscillator formula. The angular amplitude is $\theta_{0} = 6.0^{\circ} = 0.10472$ rad, so the linear amplitude along the arc is $A = L\theta_{0} = 0.9930(0.10472) = 0.10399$ m, and with $\omega = \frac{2\pi}{T} = 3.1416$ rad/s the maximum speed is $v_{max} = \omega A = 0.3267$ m/s. Second route, energy, which uses no small angle approximation at all: the bob rises $h = L(1 - \cos 6.0^{\circ}) = 0.9930(0.0054781) = 0.005440$ m at the ends of the swing, so $v = \sqrt{2gh} = \sqrt{0.10662} = 0.3265$ m/s. The two differ by $0.05$ per cent, which is exactly the size of the error the small angle approximation commits at $6^{\circ}$. Neither route needs the mass, and a solution that carries $0.250$ kg into the answer has gone wrong somewhere.

a) A closed pipe carries a displacement node at the water and an antinode at the open top, so it resonates when the acoustic length is an ODD number of quarter wavelengths: $\lambda/4$, $3\lambda/4$, $5\lambda/4$. The gap between two successive ones is therefore $\frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{\lambda}{2}$, and the end correction cancels in that subtraction, which is exactly why the experiment is done with a DIFFERENCE and not with a single length. So $\frac{\lambda}{2} = 0.494 - 0.159 = 0.335$ m, $\lambda = 0.670$ m, and $v = f\lambda = 512(0.670) = 343$ m/s.

b) $343 = 331 + 0.60\,T_{C}$ gives $T_{C} = 12/0.60 = 20$ degrees Celsius, an ordinary laboratory. This is the check that catches a slipped decimal: any speed of sound outside roughly $320$ to $360$ m/s means the wavelength was read wrong, because no teaching laboratory is at minus twenty or at plus sixty degrees.

c) Now use a single length, where the correction does not cancel. The first resonance should sit at $\lambda/4 = 0.1675$ m but was found at $0.159$ m, so $e = 0.1675 - 0.159 = 0.0085$ m, that is $0.85$ cm. With $e \approx 0.6r$ the radius is $r = 0.85/0.6 = 1.4$ cm and the tube is about $2.8$ cm across, which a ruler across the top confirms. The antinode is not exactly at the rim: the air just above the tube takes part in the oscillation, and that is the whole content of the end correction.

d) The levels are given by $L_{n} = \frac{(2n-1)\lambda}{4} - e$ with $n = 1, 2, 3$: $16.75 - 0.85 = 15.9$ cm, $50.25 - 0.85 = 49.4$ cm, and $83.75 - 0.85 = 82.9$ cm, all inside the $95$ cm tube. The fourth would need $117.25 - 0.85 = 116$ cm, longer than the tube, so three resonances and no more. A $256$ Hz fork has $\lambda = 343/256 = 1.34$ m, twice as long, so its first resonance is at $33.5 - 0.85 = 32.7$ cm and its second would need $100.5 - 0.85 = 99.7$ cm: only ONE resonance fits. Halving the frequency doubles the wavelength and therefore halves the number of resonances a given tube can show, which is why the low fork is the frustrating one in the laboratory.

a) Compare term by term with $y = A\sin(kx - \omega t)$: $A = 0.0400$ m, $k = 12.6$ rad/m, $\omega = 251$ rad/s. Then $\lambda = \frac{2\pi}{k} = \frac{6.2832}{12.6} = 0.499$ m, $f = \frac{\omega}{2\pi} = \frac{251}{6.2832} = 39.9$ Hz, $T = 1/f = 0.0250$ s, and $v = \frac{\omega}{k} = \frac{251}{12.6} = 19.9$ m/s, which also equals $f\lambda$. The sign between the two terms fixes the direction: a MINUS means the pattern moves towards positive $x$, because to keep $kx - \omega t$ constant, $x$ must grow as $t$ grows. A plus sign would have meant the other way.

b) A point of the rope does not travel with the wave; it goes up and down about its own place, in simple harmonic motion of amplitude $A$ and angular frequency $\omega$. So its greatest speed is $v_{max} = \omega A = 251(0.0400) = 10.0$ m/s, reached each time it crosses $y = 0$. That is about half the $19.9$ m/s at which the SHAPE advances, and the two numbers describe different things: one is a piece of rope moving across the rope, the other is a pattern moving along it. Nothing forbids $v_{max}$ from exceeding the wave speed, and a rope shaken harder would do it.

c) Put the numbers in: $12.6(0.200) - 251(0.0100) = 2.52 - 2.51 = 0.0100$ rad, a number of RADIANS and not of degrees, so $y = 0.0400\sin(0.0100) = 0.0400(0.0100) = 4.00\times 10^{-4}$ m, that is $+0.400$ mm, a point just leaving its rest position on the way up. A calculator left in degree mode returns $0.0400\sin(0.01^{\circ})$, roughly $7\times 10^{-6}$ m, and is wrong by a factor of $57$. The tension comes from $v = \sqrt{T/\mu}$: $T = \mu v^{2} = 0.0120(19.92)^{2} = 4.76$ N.

d) Both move, and both move in the sense that closes the gap, so both signs help: the observer moves into the waves, which raises the numerator, and the source chases its own waves, which lowers the denominator. $f' = 520\frac{343 + 20.0}{343 - 30.0} = 520\frac{363}{313} = 520(1.1597) = 603$ Hz.

e) After the crossing both motions widen the gap, so both signs reverse: $f' = 520\frac{343 - 20.0}{343 + 30.0} = 520\frac{323}{373} = 450$ Hz. The drop across the crossing is $603 - 450 = 153$ Hz, close to a musical fourth, and it happens in the moment the trains pass, which is why the pitch of a passing horn falls suddenly instead of sliding down. Why not treat the two motions separately? Moving the source alone gives $570$ Hz, moving the observer alone gives $550$ Hz, and neither the sum of the two shifts, $50 + 30 = 80$ Hz, nor anything else additive reproduces the real shift of $83$ Hz. The two effects enter as a RATIO of two factors, one in the numerator and one in the denominator, so they multiply: $570 \times \frac{363}{343} = 603$ Hz. Treating the moving source and the moving observer as interchangeable is the other classic slip, and the two numbers just computed, $570$ and $550$ Hz, show that they are not.

a) For $L_{1}$, $d_{o} = 50.0 - 35.0 = 15.0$ cm, so $d_{i} = \frac{d_{o}f}{d_{o}-f} = \frac{15.0(10.0)}{15.0-10.0} = +30.0$ cm: a real image $30.0$ cm past $L_{1}$, that is at the $80.0$ cm mark, with $m_{1} = -\frac{30.0}{15.0} = -2.00$, inverted and twice as tall. The rule that makes a two lens problem tractable is that the image formed by the first lens becomes the OBJECT of the second, real or not, and light does not care that no screen was put there. Only one thing needs watching: if that intermediate image had landed BEYOND $L_{2}$, the second lens would have a virtual object and $d_{o}$ would be negative. Here it lands before $L_{2}$, so $d_{o}$ is positive.

b) The intermediate image sits at the $80.0$ cm mark and $L_{2}$ at $100$ cm, so $d_{o2} = 20.0$ cm. Then $d_{i2} = \frac{20.0(15.0)}{20.0-15.0} = +60.0$ cm, a real image $60.0$ cm past $L_{2}$, at the $160$ cm mark, where a screen would catch it. $m_{2} = -\frac{60.0}{20.0} = -3.00$, and the total magnification is the PRODUCT, $m = m_{1}m_{2} = (-2.00)(-3.00) = +6.00$. Two inversions make an upright image: the final arrow points the same way as the original one and is $1.2(6.00) = 7.2$ cm tall. Adding the magnifications instead of multiplying them is the standard error and gives $-5$, wrong in size and in sign.

c) With $L_{2}$ at the $90.0$ cm mark, the intermediate image is unchanged at $80.0$ cm, so $d_{o2} = 10.0$ cm, which is INSIDE the focal length of $L_{2}$. Then $d_{i2} = \frac{10.0(15.0)}{10.0-15.0} = -30.0$ cm: a virtual image $30.0$ cm on the incoming side, at the $60.0$ cm mark. $m_{2} = -\frac{-30.0}{10.0} = +3.00$ and $m = -6.00$: same size as before, $7.2$ cm, but inverted and virtual, so no screen will show it. Moving a lens by ten centimetres turned a real image into a virtual one, which is the whole content of the second half of the chapter.

d) With $f_{2} = -15.0$ cm and $d_{o2} = 20.0$ cm, $d_{i2} = \frac{20.0(-15.0)}{20.0-(-15.0)} = \frac{-300}{35.0} = -8.57$ cm, a virtual image $8.57$ cm on the incoming side of $L_{2}$, at the $91.4$ cm mark. $m_{2} = -\frac{-8.57}{20.0} = +0.429$ and $m = (-2.00)(+0.429) = -0.857$, so the final arrow is inverted and $1.2(0.857) = 1.03$ cm tall, slightly smaller than the original. A diverging lens alone can never give anything but a reduced virtual upright image, but placed second in a chain it can perfectly well leave the final image inverted, because the first lens had already turned it over.

a) $\Delta y = \frac{\lambda L}{d} = \frac{(633\times 10^{-9})(3.00)}{0.250\times 10^{-3}} = 7.60\times 10^{-3}$ m, that is $7.60$ mm. Keep every length in metres before dividing: mixing nanometres and millimetres in the same fraction is the way three orders of magnitude go missing without anything looking wrong.

b) A bright fringe of order $m$ sits at $\sin\theta = m\lambda/d$, and a diffraction dark point at $\sin\theta = p\lambda/a$. When the two angles coincide the interference maximum falls exactly where one slit alone sends no light at all, so nothing arrives and the fringe is MISSING. Setting $\frac{m\lambda}{d} = \frac{p\lambda}{a}$ gives $m = p\frac{d}{a} = 5p$, so $m = 5, 10, 15,\ldots$ are missing on each side. The ratio $d/a = 5$ is a whole number here, which is why the effect is clean; with $d/a = 4.3$ no order would vanish completely.

c) The first diffraction minimum is at $\sin\theta = \frac{\lambda}{a} = \frac{633\times 10^{-9}}{0.0500\times 10^{-3}} = 0.01266$, so $y = L\sin\theta = 3.00(0.01266) = 0.0380$ m $= 38.0$ mm on each side, and the central envelope is $76.0$ mm wide. Inside it the fringes are $7.60$ mm apart, and $38.0/7.60 = 5$ exactly, which says that the missing order $m = 5$ is the envelope edge itself. What survives is the central fringe plus orders $1$ to $4$ on each side: $9$ bright fringes, of quickly fading brightness, as the solution figure shows. Swapping the roles of $d$ and $a$ in these two formulas is the single most common mistake of the chapter: the SMALL opening, $a$, controls the big feature, and the big spacing, $d$, controls the fine one.

d) Covering one slit removes the interference, because interference needs two beams to add. The regular $7.60$ mm fringes vanish. What stays is the diffraction pattern of the surviving slit alone, whose central bright band is still $76.0$ mm wide, because that width depends on $a$ and on nothing else, and $a$ has not changed. The screen also becomes much darker, since half the light is now blocked and the bright fringes were four times the brightness of one slit rather than twice. The experiment therefore separates cleanly: $d$ owns the fringes, $a$ owns the envelope.