Helpppp plz. EM Radiation doesn't make sense to me

premed

Hi. I don't see how you would get the answers to these questions.
1) A beam of light is shone on two sheets of paper, one of which is perfectly absorbing black and the other perfectly reflecting white. If the radiation pressure on the white paper is "p", the radiation pressure on the black paper will be ....
p/2. How is this the answer? What equation would you use and what does it mean conceptually?
2) A ray of light follows the path shown as it reaches the boundary between two transparent materials( the ray bounces off and reflects at the point of incidence. It stays in the material with index of refraction n1 and doesn't go into material with index n2). What could you conclude about the relative indexes of refraction of these two materials? the answer is n1>n2 but i don't get why. If the ray would travel faster in material 2 (n=(c/v), why would it reflect off rather than refracting through it? Please explain why it reflects both conceptually and quantitavely if possible. Thx

LtStorm

For your first question I want to say the reason is because light hitting the white reflective surface exerts pressure, then the light bouncing away exerts pressure again. In contrast, the light exerts pressure on the black absorptive surface, but since it's absorbed by the surface it doesn't push against it twice like when reflected. So, it's pressure would be p/2 of the white surface.

Shooting Star

You have to read up on "total internal of reflection of light". After that, if there are questions, you can ask here any time.

premed

But i don't understand WHY it exerts pressure twice when it reflects. If a photon hits an object, doesn't it only exert pressure as it bounces off? Isn't the enitre collision only one instant? Tell me if this analogy works cuz it kinda helps me to understand but I'm not sure if it is applicable: Is it like I am jumping off a platform onto the ground and then immediately jump back up? Am I like a photon exerting pressure as I jump onto the ground and again as I push off?

mikelepore

Suppose a ball has momentum p1 = +X, hits a wall, bounces back, it's new momentum is p2 = -X , negative, same magnitude but opposite direction. The change in momentum of the ball is p2-p1 = (-X)-(+X) = -2X.

kdv

Yes, the pressure is exerted only once (at the instant of the collision) no matter if the photon is absorbed or if it is reflected. The point is rather that when the photon is bounced back, twice the amount of momentum is transferred to the surface (but this transfer only occurs once, during the collision).

Total Momentum is conserved in both cases but the key point is that the object being hit will have twice the initial momentum of the photon if the photon bounces back.

tiny-tim

Hi premed!

(Like mikelepore's example, but with a movable target:)

Suppose you're stationary on ice, and someone throws a ball at you, and you catch it. How much momentum does that give you?

Now suppose instead that you don't see it coming, and it hits your chest (at the same speed) and bounces off you. How much momentum does that give you?

premed

awwww ok i get it. its momentum is first positive going towards the surface then negative when leaving the surface, which means its magnitude of change is 2x rather than x. Thanks a lot.

LtStorm

Yeah, the surface absorbs the photon, but re-emits it if it's a reflective white surface.

jswtp

Using the thrown ball analogy: Photons hitting a black surface are like a ball hitting a soft, padded surface (like a down jacket): The jacket absorbs some of the impact, so not all of the pressure is exerted against the person wearing the jacket. But photons hitting a white surface are like a ball hitting a rigid material -- the ball bounces back, and exerts greater pressure.