34: Problem 34.45

INTRO:
The intensity of sunlight reaching the earth is 1360 W/m².

Part A:
What is the power output of the sun?
Express your answer with the appropriate units.

SOLUTION:
givens:
I = 1360 W/m²

By rearranging the Intensity formula we are able to solve, since I = P_source⋅1/(4πr²),
P_source = I⋅4πr²
r is the distance from the sun to earth and that's something we're going to have to google
r = 150,000,000 km = 1.5⋅10⁸km ⋅1⋅10³m/km = 1.5⋅10¹¹m
& r²=2.25⋅10²²m²
→P_source = 1360 W/m²⋅4π⋅2.25⋅10²²m²
=1360⋅4π⋅2.25⋅10²² W
P_source = 3.85⋅10²⁶ W

Part B:
What is the intensity of sunlight on Mars?
Express your answer with the appropriate units.

SOLUTION:
first google the distance between the sun and mars to get r
r = 2.28⋅10¹¹ m & r² = 5.20⋅10²² m²

& plugging into the formula we get
I = 3.85⋅10²⁶ W ⋅1/(4π⋅5.20⋅10²² m²)
≈ 598 W/m²

34: Problem 34.22

Part A:
At what distance from a 15W point source of electromagnetic waves is the magnetic field amplitude 0.60μT ?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
givens:
P_source = 15 W
B₀ = 0.60μT
and we're trying to determine the distance from the point source, r

How can we relate the magnetic field amplitude and the distance from the point source? 
neither of our basic field formulas helps us, but we do know a series of equations relating the Electric field amplitude and electromagnetic wave intensity, and the amplitudes of the Electric field and the Magnetic field are related by the speed of light, we can arrange the following formulas to solve for r

Intensity relations~ 
I= P_source⋅1/(4πr²) = c⋅ε₀⋅E₀²⋅1/2
Magnetic & Electric field relations~
B₀⋅c = E₀
so E₀² = (B₀⋅c)² 
∴ I= c⋅ε₀⋅(B₀⋅c)²⋅1/2=c³⋅ε₀⋅(B₀)²⋅1/2 
since I also equals P_source⋅1/(4πr²), it can be simplified that
P_source⋅1/(4πr²) = c³⋅ε₀⋅(B₀)²⋅1/2
now, ε₀ = 1/(μ₀c²) = 1/[(4π⋅10^-7 H/m)c²]
so P_source⋅1/(4πr²) = c³⋅1/[(4π⋅10^-7 H/m)c²]⋅(B₀)²⋅1/2 simplifies to
P_source⋅[2⋅10^-7 H/m] / (r²) = c⋅(B₀)² which simplifies to
r²=P_source⋅(2⋅10^-7 H) ⋅1/[c⋅B₀² ⋅m]
leaving r = √(P_source⋅(2⋅10^-7 H) ⋅1/[c⋅B₀² ⋅m])
now, plugging in for all of the values for which we've arranged so nicely, we get
r = √{(15 W)⋅(2⋅10^-7 H)⋅1/[(3⋅10⁸ m/s)(3.6⋅10^-13)⋅T)² ⋅m] }
→ the first thing I'm going to do here is solve for the units & check that I'm on the right path
√{W⋅H⋅1/(m/s)⋅1/(T²)⋅1/m}
W = kg⋅m⋅m⋅1/s⋅1/s⋅1/s & H = kg⋅m⋅m⋅1/s⋅1/s⋅1/A⋅1/A & T = kg⋅1/A⋅1/s⋅1/s →1/T = A⋅s⋅s⋅1/kg
⇒√{kg⋅m⋅m⋅1/s⋅1/s⋅1/s⋅kg⋅m⋅m⋅1/s⋅1/s⋅1/A⋅1/A⋅A⋅s⋅s⋅1/kg⋅A⋅s⋅s⋅1/kg⋅1/m⋅s⋅1/m}
⇒√{kg²⋅m⁴⋅A²⋅s⁵⋅1/kg²⋅1/m²⋅1/s⁵⋅1/A²}⇒√{m²}
⇒units=m
which is right so that means our equation is set up properly, now solve
√{(15)⋅(2⋅10^-7)⋅1/[1.08⋅10^-4] = √{1/36} = 1/6 = 0.17 m

22: Problem 22.18

INTRO:
Light of 630nm wavelength illuminates a single slit. The intensity pattern shown in the figure is seen on a screen 2.2m behind the slits.

Part A:
What is the width (in mm) of the slit?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
givens: 
λ = 630 nm
L = 2.2 m

looking at the Intensity pattern, its clear that the y₁ dark fringes occur at x=1cm and x=2cm so the width of the maxima, is 1 cm
w = 1 cm

Now by rearranging the formula w=2λL⋅1/a, an equation can be formed to solve for the slit width a
a = 2λL⋅1/w = 2⋅630 nm⋅2.2 m⋅1 / 1 cm = 0.28 mm

22: PhET Tutorial: Wave Interference

Part A:
How does the amplitude of the wave depend on the distance from the source?

A) The amplitude decreases with distance.
B) The amplitude increases with distance.
C) The amplitude is constant.

SOLUTION:
Comparing the reading from the detector placed close to the source with a reading placed farther away from the source, one can see that the amplitude decreases as distance increases. Option A)

Part B:
Which statement best describes how the intensity of the wave depends on position along the screen?

A) The intensity is roughly constant.
B) The intensity is large near the middle of the screen, then decreases to nearly zero, and then increases again as the distance from the middle of the screen increases.
C) The intensity is a maximum near the middle of the screen (directly to the right of the source) and significantly decreases above and below the middle of the screen.

SOLUTION:
The intensity graph appears to be a vertical line, or pretty much constant all along the screen. So it can be assumed that the intensity is roughly constant, or option A)

Part C:
Which statement best describes how the intensity of the wave depends on position along the screen?

A) The intensity is large near the middle of the screen, then decreases to nearly zero, and then increases again as the distance from the middle of the screen increases.
B) The intensity is roughly constant.
C) The intensity is a maximum near the middle of the screen (directly to the right of the source) and significantly decreases above and below the middle of the screen.

SOLUTION:
Just as in the previous question, the intensity graph appears to be a vertical line, or roughly constant. Option A)

Part D:
Which statement best describes how the intensity of light on the screen behaves?

A) The intensity is large near the middle of the screen, then decreases to nearly zero, and then increases again as the distance from the middle of the screen increases.
B) The intensity is roughly constant.
C) The intensity is a maximum near the middle of the screen (directly to the right of the source) and significantly decreases above and below the middle of the screen.

SOLUTION:
The intensity graph appears to be large in the middle of the screen, drop down to zero and then increase again as distance from the middle increases. This is option A)

Part E:
How do the distances r1 and r2 compare?

A) The difference in the distances is equal to half the wavelength of the wave.
B) The difference in the distances is equal to the wavelength of the wave.
C) The difference in the distances is equal to a quarter of the wavelength of the wave.
D) The distances are the same.

SOLUTION:
my measured wavelength of green light ~ 600 nm
r1 = 2790.33 nm
r2 = 3364.33 nm
r2-r1 = 574 nm
this shows that the difference in the distances is roughly equal to the wavelength of the wave, or option B)

Part F:
Compare the distances from the first location nearest the middle of the screen where the intensity is nearly zero (dark fringe) to each of the two slits. How do the distances compare?

A) The difference in the distances is equal to a quarter of the wavelength of the wave.
B) The distances are the same.
C) The difference in the distances is equal to half the wavelength of the wave.
D) The difference in the distances is equal to the wavelength of the wave.

SOLUTION:
r1 = 2807.86 nm
r2 = 3160.03 nm
r2 - r1 = 352.17 nm
the difference in distances is roughly equal to half the measured wavelength, or option C)

Part G:
How does the distance between consecutive bright fringes depend on the wavelength of the light?

A) The spacing of the fringes does not change when the wavelength changes.
B) The fringes get closer together as the wavelength increases.
C) The fringes get farther apart as wavelength increases.

SOLUTION:
distance btwn bright fringes w/ green wavelength- 418.04 nm
distance btwn bright fringes w/ blue wavelength- 243.72 nm

since blue has a shorter wavelength than green and blue's distance btwn is less than greens, it can be said that either 
distance between bright fringes decreases as wavelength is decreased
or distance between bright fringes increases as wavelength is increased
the second one satisfies option C)

Part H:
How does the distance between consecutive bright fringes depend on the slit separation?

A) The fringes get farther apart as the slit separation increases.
B) The fringes get closer together as the slit separation increases.
C) The spacing of the fringes does not change when the slit separation changes (just the brightness changes).

SOLUTION:
(I used yellow light)
distance btwn bright fringes w/ 875 slit separation- 556.94 nm
distance btwn bright fringes w/ 1750 slit separation- 290.11 nm
so the lesser separation resulted in a greater distance, therefore it can be said either
the distance between the bright fringes increases as slit separation decreases 
or the distance between the bring fringes decreases as slit separation increases
the second satisfies option B)

PART I:
How does the distance between the bright fringes depend on the slit width (for slit widths less than the wavelength of the light)?

A) The spacing of the fringes does not change when the slit width changes.
B) The fringes get closer together as the slit width increases.
C) The fringes get farther apart as the slit width increases.

SOLUTION:
for ~ 700 nm wavelength
distance btwn bright fringes w/ notch 1 slit width(~197.32 nm)- 197.32 nm
distance btwn bright fringes w/ notch 2 slit width(~498.93 nm)- 200.31 nm
so it can be said that the spacing of the fringes does not change as the slit width changes for slit widths less than the wavelength of the light, or option A)

Part J:
How does the distance between the bright fringes depend on the amplitude of the wave?

A) The fringes get farther apart as the amplitude increases.
B) The spacing of the fringes does not change when the amplitude changes (just the brightness changes).
C) The fringes get closer together as the amplitude increases.

SOLUTION:
Option B)

Part K:
Does interference occur when water or sound waves encounter a barrier with two slits?

A) Interference also occurs for sound waves, but not for water waves.
B) Interference also occurs for water waves, but not for sound waves.
C) Yes, interference also occurs for both of these types of waves.
D) No, it only occurs for light.

SOLUTION:
Interference is a very common phenomenon that can occur with any type of wave.
Option C)

22: Two-Slit Interference

INTRO:
As Richard Feynman stated in his book on quantum mechanics,

Interference contains the heart and soul of quantum mechanics.

In fact, interference is a phenomenon of classical waves, easily perceived with sound or light waves. (It contains the soul of quantum mechanics only after you swallow the preposterous notion that particles in motion are described by a wave equation rather than the laws of Newtonian mechanics.)

In this problem, you will look at a classic wave interference problem involving electromagnetic waves. Young's double-slit experiment provided an irrefutable demonstration of the wave nature of light and is certainly one of the most elegant experiments in physics (because it demonstrates the important concept of interference so simply). For the purposes of this problem, we assume that two long parallel slits extending along the z axis (out of the plane shown in the figure) are separted by a distance d. They are illuminated coherently, that is, in phase, by light with a wavelength λ, for example by a laser beam polarized in the z direction. (Lacking a laser, Young used an intense source diffracted by a slit to produce coherent illumination of his double slits.)

The key point is that the electric field far downstream from the slit (e.g., at a large positive x value) is the sum of the electric fields emanating from each of the two slits. Hence, the relative phase of these electric fields at some observation point O determines whether they add in phase (constructively) or out or phase (destructively).(Figure 1)

To refresh your memory about traveling waves, the electric field E(x,t) that is incident on the double slits from the left is a function of x and t. Let us assume that it has amplitude Eleft. We will also assume a cosine trigonometric function with the arbitrary phase set equal to zero (i.e., at the point x=0, t=0 you find that E=E_left). Then
E(x,t)=E_leftcos[2π/λ (x−ct)].

The argument of the cosine function is the phase Φ(x,t). It can be written as Φ(x,t)=kx−ωt, where ω is the angular frequency (ω=2πf) and k is the wave number defined by k=2π/λ. The phase increases by 2π each time the distance increases by λ. Moreover, the phase is constant for an observer moving in the positive x direction at the speed of light, i.e., for whom x = ct.

Figure 1

Figure 2

PART A:
(Figure 2) Now consider the electric field observed at a point O that is far from the two slits, say at a distance r from the midpoint of the segment connecting the slits, at an angle θ from the x axis. Here, far means that r≫d, a regime sometimes called Fraunhofer diffraction.

The critical point is that the distances from the slits to point O are not equal; hence the waves will be out of phase due to the longer distance traveled by the wave from one slit relative to the other. Calculate the phase Φ_lower(O,t) of the wave from the lower slit that arrives at point O.

Express your answer in terms of d, θ, λ, c, r, t, and constants like π.

SOLUTION:
Φ_lower(O, t) = 2π/λ (r−ct) + π/λ dsin(θ)

Part B:
Now calculate the phase Φ_upper(O,t) of the wave from the upper slit that arrives at point O.
Express your answer in terms of d, θ, λ, c, r, t, and constants such as π.

SOLUTION:
Φupper(O, t) = 2π/λ (r−ct) - π/λ dsin(θ)

NOTE:
In order to make the math as simple as possible, we will define two phases:
ϕ = 2π/λ (r−ct) and δϕ=π/λ dsin(θ).
Then Φ_lower= ϕ + δϕ and Φ_upper= ϕ − δϕ.

Part C:
Assuming that the maximum amplitude of the field at point O for a wave from midway between the slits is E(r), now find the magnitude of the combined field E at O due to the two slits. You may ignore variations in the maximum amplitude and consider only variations in phase of the waves emerging from the slits.
Express your answer in terms of E(r), ϕ, and δϕ.

SOLUTION:
so, begin by finding both of the magnitudes of the electric field due to the slits
since E(x,t)=E_leftcos[Φ(x,t)] & Φlower(O, t) = ϕ + δϕ
Elower(O, t) = E(r) cos[ϕ + δϕ]
& with Φupper(O, t) = ϕ - δϕ,
Eupper(O, t) = E(r) cos[ϕ - δϕ]
To get the magnitude of the combined field at O, simply add the two separate fields
E = E(r) cos[ϕ + δϕ] + E(r) cos[ϕ - δϕ]
= E(r)[cos(ϕ + δϕ)+cos(ϕ - δϕ)]
If we use the identity cos(A+B)+cos(A-B)=2cos(A)cos(B),
we're able to fully simplify the answer to E = 2E(r)cos(ϕ)cos(δϕ)

Part D:
The key aspect of two-slit interference is the dependence of the total intensity at point O on the angle θ. Find this intensity I(θ).
The formula for intensity is

I=ϵ₀c(amplitude of E)²⋅½.

Express your answer in terms of I_max, θ, d, and λ, where I_max=2ε₀cE(r)². Note: cos²x should be coded as cos(x)^2.

SOLUTION:
Recall that E can be written as the product of the amplitude and the cosine of the phase, where the phase depends on time. In the above expression for E, the cos(ϕ) term is a function of time, whereas the rest of the variables/functions are not. Therefore, the amplitude of E is 2E(r)cos(δϕ). 
To find the dependence of I on θ, recall that δϕ = π/λ dsin(θ) and substitute for that in the equation
this results in 2E(r)cos(π/λ dsin(θ))

we have a formula in the book that states that I_{double slit}=4I_maxcos²[πdy⋅1/(λL)] and also one that states I_{double slit}=4I_maxcos²(θ)

using these we are able to determine that I(θ) = I_maxcos²(π/λ dsinθ)

Part E:
Two-slit interference is usually observed at small angles, and thus sin(θ) can be replaced by just θ. In this limit, the important observable is the spacing between successive minima (or maxima) of the interference pattern. Find the angular spacing Δθ of the interference pattern.
Express your answer in terms of d, λ, and any needed constants.

SOLUTION:
The intensity has the form Imaxcos2(π/λ dsinθ)
Since this depends on a cosine squared, the phase difference between two maxima or minima is π, not 2π, as it would be for a simple cosine function. Thus, to find the angular separation, you must set the difference in phase between two different directions equal to π:
(π/λ dsinθ₂)−(π/λ dsinθ₁)=π.
so
(πdsinθ₂)−(πdsinθ₁)=πλ
→dsinθ₂−dsinθ₁=λ
→sinθ₂−sinθ₁=λ/d

Note that once you make the substitution sin(θ)=θ for both phases, you will have a relatively simple expression involving Δθ, where Δθ=θ₂−θ₁.
sinθ₂−sinθ₁=λ/d 
→θ₂−θ₁=λ/d 
Δθ=λ/d