Comments on Profile Post by Fluffinya

depends what level, high school or college ?

anyway i'll say everything i know..
so as for the exact reason of why light is diffused when going through an aperture with the dimension near its wavelength, it is due to a phenomenon analogous to refraction, because when light go through an atom, there is a difference in its environment that will deviate its trajectory.

however, i would not be able to explain you in details how it works for now as i'm not well versed in this subject.

now, for the interference figures, it is due to the superposition principle that allows you to sum the value of 2 wave.

however, just saying that won't help much. thus, let's define the physical properties of a wave, a wave is dependant of time and space, what i mean by that is that its value (in volt) will change depending on where it is and when you mesure.

thus we will determine the value of our wave so:
A(x,t)=Amax*cos(w*t+k*x+P) or
A(x,t)=Amax*exp(i(w*t+k*x+P))
where
Amax is the Amplitude of the wave
w its wavepulse (w = f/2*pi)
k its wave vector (k = c/w)
P the origin phase
c the celerity of the wave

now let's go back to what i said earlier: using the superposition principle, we add two wavelength. to simplify the equation, let's say that we take x=0 and P=0. thus we have 2 waves A1 and A2 of pulsation w1 and w2 respectively

omo thank u

i finally understood

what did you not understand btw ?

i skipped classes the day when tchr taught diffraction.

well if i can give you a tip, try to learn by hearth the relations as soon as possible as to not forget them, the rest, it's just trigonometry and logic~

I'll try my best i'm good in math or physics

btw if you wanna hear my explanation on interference, i'm only halfway through

If you want to ^_^

so from where i left, using the superposition principle and x=0:
Atot(0,t)= A1*cos(w1*t + P1) + A2*cos(w2*t + P2)
however, our eyes cannot perceive the variation of light over time but rather its average value, it's average value being sqrt(<Atot²>)
sqrt: square root
< .. >: the integer from 0 to T of a function divided by T with T=2pi/w1 or w2

the reason i use Atot² in the < .. > instead of Atot is because we need the absolute value of the integer as otherwise, <Atot>=0, it's a sinusoidal function after all !!
thus, we get :
Atot(0,t)²=A1²*cos²(w1*t + P1) + A2²*cos²(w2*t + P2) + A1*A2*cos(w2*t + P2)*cos(w1*t + P1)
we have <cos²>=1/2 and:
cos(a)*cos(b) = (1/2)*(cos(a+b)+cos(a-b))

thus we get:
A1*A2*cos(w2*t + P2)*cos(w1*t + P1) = 0.5*A1*A2*(cos((w1 + w2)*t +P1+P2) + cos((w1 - w2)*t +P1-P2)
thus, we get <cos((w1 + w2)*t +P1+P2)>=0 no matter what.
however, if w1=w2, we have cos((w1 - w2)*t +P1-P2) = cos(P1-P2)
and since it is constant over time, <cos(P1-P2)> = cos(P1-P2)
thus we get :
<Atot(0,t)²>=(A1² + A2²)/2 + A1*A2cos (P1-P2)

now we will simplify the equation by saying A1=A2=A and P1-P2 = Pd
thus we get:
<Atot(0,t)²>= A(1 + cos(Pd)) where Pd is the phase difference between the 2 waves, it s basically what you get when you diffract light coming from the same lightsource using 2 apertures, in fact, since it comes from the same lightsource, they both have the same wavelength, thus w1=w2 is repected

and then since the do not take the same route to a given point, they also do not have the same initial phase, thus the phase difference will vary depending on what point you take in the screen.

now, how do we find the fringe spacing?
for that, we'll need to find the difference P=P1-P2.
to find that value, we'll need to find the difference in traveled distance between the 2 waves.
so we'll need to introduce new value:
a: the spacing between the 2 apertures
L: the distance between the apertures and the screen
x: the vertical coordinate of any given point on the screen

now, we assume that the line that goes through the 2 apertures is parallel to the screen, we take the origin of our frame of reference between the 2 apertures and orient it as you please.
we now need to find the difference in distance traveled to find the phase difference between the two waves.

so for the wave that comes from the aperture below, the distance traveled is:
on the X axis, it is x+a/2
on the Y axis, it is L
to find the distance traveled, we use Pythagoras theorem:
D = sqrt( X² + Y²)
thus we get:
D1 = sqrt( (x +a/2)² + L²) then, for the other aperture:
D2 = sqrt( (x- a/2)² + L²)

we now have to find the difference between the 2 values, for that, we'll have to do a Taylor expansion of the 2 values:
D1 = L* (1 + ((2*x + a) / (2*L))² )**(1/2) ~~ 1 + L*(2*x + a)² / (2*L)²*(1/2)
we now have:
D1 = 1 + (4*x²) / (4*L) + (2*x*a / 4*L) + ( a/ 2*L)²
or, ( a/ 2*L)² << 1 thus,
D1 ~= 1 + (4*x²) / (4*L) + (2*x*a / 4*L²)

D2 = L* (1 + ((2*x - a) / (2*L))² )**(1/2) ~~ 1 + L*(2*x - a)² / (2*L)²*(1/2) thus:
D2 ~= 1 + (4*x²) / (4*L) - (2*x*a / 4*L)
we now need to find the difference between the 2 values:
D1 - D2 = 1 + (4*x²) / (4*L) + (2*x*a / 4*L²) + ( a/ 4*L) - (1 + (4*x²) / (4*L) - (2*x*a / 4*L))
D1 - D2 = (x*a) / L

now let's see the relations we have:
w = f / 2*pi
f = c / lambda
thus to find the relation between the phase difference and the distance traveled, we do:
((D1-D2) / lambda) * 2 * pi = phase difference
... well the relations earlier are kinda useless

so Pd = x*a*2*pi / L*lambda
and the distance between two luminous fringe is the distance needed for cos(Pd) to be equal to 1 so:
cos(Pd) = 1
<=> Pd = 2*k*pi with k a real number
<=> x*a*2*pi / L*lambda = 2*k*pi
<=> x = L* k * lambda / a
<=> x =L* lambda / a for the difference between 2 fringes

also, the <=> symbol is the symbol for the equivalent in france i don't know if you have the same where you are.

so.. that concludes my demonstration, now if you want to impress your physics teacher, you can learn that by hearth and show it to him XD
because, well i did write this without even looking at my lesson *proud*

and if you don't understand everything don't hesitate to ask me some questions as i did not explain every details ( notably the Taylor expansion, which is a math problem) and i know it can be confusing for someone who's never seen it beforehand

she'll be impressed because the only thing im good at is memorizing

try to make a sketch of the situation btw but i think it's the same than the one you see in class.