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If an object is illuminated by a point source of light, then the illuminance at the object is equal to the luminous flux of the light source, divided by the surface area of the sphere, whose radius is equal to the distance the object is from the light source



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Answer (for Objective 51): E and F are correct E is correct because, as it stands, Bicycle accesses Wheel only through its informal API, the method and the constructor F is correct because Wheel is not well encapsulated, and making size private would make Wheel well encapsulated A is incorrect because the code is legal B and C are incorrect because, in general, encapsulation is mostly independent of cohesion, and Bicycle is already treating Wheel in a loosely coupled way D is incorrect based on the above





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Remember that the luminous flux of the light source is spreading out spherically in all directions, so only a fraction of the luminous flux is available to illuminate the desk Use of this equation is valid only if the light from the luminous source strikes perpendicular to the surface it is illuminating It is also only valid for luminous sources that are small enough or far enough away to be considered point sources Thus, the equation does not give accurate values of illuminance for long, fluorescent lamps or incandescent lightbulbs that are close to the surfaces that they illuminate

If z is constant, then y is directly proportional to x When x increases, y increases When x decreases, y decreases If x is constant, then y is inversely proportional to z 2 When z 2 increases, y decreases When z 2 decreases, y increases

If r is constant, then E is directly proportional to P When P increases, E increases When P decreases, E decreases If P is constant, then E is inversely proportional to r 2 When r 2 increases, E decreases When r 2 decreases, E increases

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Illumination of a Surface What is the illuminance at on your desktop if it is lighted by a 1750-lm lamp that is 250 m above your desk

20 Given:

inversion region, it increases) The inversion layer forms a duct that acts like a waveguide In Fig 2-19, the distance Dl is the normal radio horizon distance, and D2 is the distance over which duct communications can occur Ducting allows long-distance communications from lower VHF through microwave frequencies, with 50 MHz being a lower practical limit and 10 GHz being an ill-defined upper limit Airborne operators of radio, radar, and other electronic equipment can sometimes note ducting at even higher microwave frequencies Antenna placement is critical for ducting propagation Both the receiving and transmitting antennas must be either (1) inside the duct physically (as in airborne cases) or (2) able to propagate at an angle such that the signal gets trapped inside the duct The latter is a function of antenna radiation angle Distances up to 2500 mi

Assume that the lightbulb is the point source Diagram the position of the bulb and desktop Label P and r Known:

250 m

Unknown:

2 public class Mouthwash { 3 static int x = 1; 4 public static void main(String[] args) { 5 int x = 2; 6 for(int i=0; i< 3; i++) { 7 if(i==1) Systemoutprint(x + " "); 8 } 9 go(); 10 Systemoutprint(x + " " + i); 11 } 12 static void go() { int x = 3; } 13 }

The surface is perpendicular to the direction in which the light ray is traveling, so you can use the point source illuminance equation E

P 4 r2 175 103 m 4 (250 m)2

250 m

or so are possible through ducting Certain paths where frequent ducting occurs have been identified: the Great Lakes to the Atlantic seaboard; Newfoundland to the Canary Islands; across the Gulf of Mexico from Florida to Texas; Newfoundland to the Carolinas; California to Hawaii; and Ascension Island to Brazil Another condition is noted in the polar regions, where colder air from the land mass flows out over warmer seas (Fig 2-20) Called subrefraction, this phenomena bends EM waves away from the earth s surface thereby reducing the radio horizon by about 30 to 40 percent

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