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Assuming that 10.0% of a 100-W light bulb’s energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, and that the photons spread out uniformly and are not absorbed by the atmosphere, how far away would you be if 500 photons per second enter the 3.00-mm diameter pupil of your eye? (This number easily stimulates the retina.)

181 $\mathrm{km}$

Physics 103

Chapter 29

Introduction to Quantum Physics

Quantum Physics

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

University of Winnipeg

Lectures

02:51

In physics, wave optics is…

10:02

Interference is a phenomen…

03:37

Assuming that 70% of the E…

01:52

Defend the statement that …

01:33

Suppose that 25% of people…

01:26

If 10% of a 50-kg rock cli…

00:50

Evaluate.$$\left(\begi…

00:17

Evaluate.$$100(72.…

00:16

Evaluate.$$6504 \d…

00:15

Divide.$$-100 \div 10$…

01:05

Multiply. $$(0.1508)(-100)…

01:30

Evaluate each expression.<…

02:10

If an automobile were to h…

02:12

Suppose that a 5% increase…

01:42

If a 10% decrease in the p…

00:34

Fill in the blank with the…

00:33

Multiply: $100(0.67 x)$

00:55

Divide. Then check by mult…

00:25

Evaluate the expression.

00:41

Evaluate.$$99 \% \…

02:35

Evaluate.$$-|10|

So in this problem, we have a long lighting warm. All right. And, uh, so we assume that ah ah, in this ah, lighting bump. Only 10% off the total. Ah, emitted emitted. The photos has, uh they're in the visible visible wrench and their average wait glances of 5 80 centimeters and total power off this landing. Bobby is 100 watts. So, uh, we want to figure out Ah ah, basically. So there's a lot of hope for that. You need to stand away so that you can still see this line in bulk. Okay, so according to this set up, we can fit. We can find out the energy off this. Ah, visible full time so that it would be be equal h time. See you, Fernanda. All right, so we already have the Lunda, so we can figure this Very Reza 3.43 times. 10 to the negative. 19. Jules. All right. And because we have the power of the lining bomb, which is 100 volts. So each second the energy emitted is 100. Joe's, right. So the total number Ah, full time I made it out. His, uh, and a quote P Times, T Ory, Proteas one second and P s 100 watts and e is over here. Okay, so you can see that this value is ah, 2.92 times turned to the twenties. So that means h second, uh, there's ah total. There's totally, uh, 2.9 and two times 10 2 twenties photos come comes out. But ah, according to this problem on in, 10% of them will be in the visible visible range. So let's say in visible equal end times 10%. Right, So this gives you two point and two times 10 to the 19th. So, uh, we already have this invisible. So the model is S o. Here's the lightning. Bolt it, Shut it. Shut the light. Ah, really? So in that case, it is like a spherical surface, right? So the radius of the spirits off of this theories are So what we have in visible there are four pi r squared, which is the surface off this for on this fear equal should be equal to the limit. The extreme density. So which is 500 according to this problem, way have 500 photons in the, uh, in the area, uh, whose diameter is three millimeters. So areas pie 23 millimeters squared. Right. So you can sew for our And in this equation only only verbally. This arm, because invisible is ah is over here. So you can basically find that are equal 1 81 que meters. All right, So that means even if you stand ah, in the position which is a 1 81 kilometers away from the night Melanie month, you can steal situs. Ah, you can still see this lining good.

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