34: Electromagnetic Waves Ranking Task

Part A:
Rank these electromagnetic waves on the basis of their speed (in vacuum).
Rank from fastest to slowest. To rank items as equivalent, overlap them.

AM radio wave, yellow light, X-ray, infrared light, green light, FM radio wave

SOLUTION:
all the light waves have the same speed, c; The speed of light in a vacuum
v_{AM radio} = v_{FM radio} = v_infrared = v_yellow = v_green = v_X-ray 

Part B:
Rank these electromagnetic waves on the basis of their wavelength.
Rank from longest to shortest. To rank items as equivalent, overlap them.

AM radio wave, yellow light, X-ray, infrared light, green light, FM radio wave

SOLUTION:
looking at a spectrum displaying the varying wavelengths, one can rank the waves accordingly

λ_{AM radio} ≥ λ_{FM radio} ≥ λ_infrared ≥ λ_yellow ≥ λ_green ≥ λ_X-ray

Part C:
Rank these electromagnetic waves on the basis of their frequency.
Rank from largest to smallest. To rank items as equivalent, overlap them.

AM radio wave, yellow light, X-ray, infrared light, green light, FM radio wave

SOLUTION:
looking at a spectrum displaying the varying frequencies, one can rank the waves accordingly
f_{AM radio} ≤ f_{FM radio} ≤ f_infrared ≤ f_yellow ≤ f_green ≤ f_X-ray

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: Problem 22.12

INTRO:
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm(red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 510lines/mm , and the light is observed on a screen 1.3m behind the grating.

Part A:
What is the distance between the first-order red and blue fringes?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
givens: 
λ_red = 656 nm
λ_blue = 486 nm
line density = 510 / mm
L = 1.3 m
d = 1 mm / 510 lines = 1.961⋅10^-3 mm
m = 1

the positions can be determined by using the formula θ=sin^-1(mλ/d) & Ltan(θ)=y
θ_red = sin^-1(1⋅λ_red/d) = sin^-1(656 nm / 1.961⋅10^-3 mm)
θ_red = 0.341099 rads = 19.5435°
y_red=Ltan(θ_red) = (1.3 m)tan(19.5435°) = 0.462 m
θ_blue = sin^-1(1⋅λ_blue/d) = sin^-1(486 nm / 1.961⋅10^-3 mm) 
θ_blue = 0.250443 rads = 14.349°
y_blue=Ltan(θ_blue) = (1.3 m)tan(14.349°) = 0.333 m

so the blue fringe begins at 0.333 m and the red fringe begins at 0.462 m
the distance in between the fringes is y_red-y_blue = 0.129 m = 13 cm

22: Problem 22.9

INTRO:
A 4.0-cm-wide diffraction grating has 2000 slits. It is illuminated by light of wavelength 590nm .

Part A:
What is the angle (in degrees) of the first diffraction order?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
givens: 
width = 4 cm
d = 4 cm / 2002 spaces between slits = 1.998⋅10^-3 cm
N = 2000
λ = 590 nm
m = 1

the angle can be determined by using the formula θ=m⋅λ/d
θ= 590 nm / 1.998⋅10^-3 cm
θ= 0.02953 radians = 1.69°

Part B:
What is the angle (in degrees) of the second diffraction order?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
do the same thing but with m=2
θ=2⋅590 nm / 1.998⋅10^-3 cm
θ= 0.05906 radians = 3.38°

22: Problem 22.7

INTRO:
In a double-slit experiment, the slit separation is 200 times the wavelength of the light.

Part A:
What is the angular separation (in degrees) between two adjacent bright fringes?
Express your answer with the appropriate units.

SOLUTION:
givens: 
d = 200λ

if we consider wavelength as a variable (such as x or y) for a moment, we can just plug the given for d directly into the formula Δθ=λ/d and get λ/200λ which allows us to simply cancel out the λ
giving us Δθ = 1/200 which is the angular separation in radians which is 0.286°

22: Problem 22.6

INTRO:
Light from a sodium lamp (λ=589nm)illuminates two narrow slits. The fringe spacing on a screen 110 cm behind the slits is 3.6mm.

Part A:
What is the spacing (in mm) between the two slits?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
givens: 
λ = 589 nm = 5.89⋅10^-7 m
L = 110 cm = 1.1 m
Δy = 3.6 mm = 3.6⋅10^-3 m

we can rearrange the formula Δy = λ⋅L/d to give us the distance between the slits, d
d=λ⋅L/Δy → d = (5.89⋅10^-7 m)⋅(1.1 m/3.6⋅10^-3 m) = 1.8⋅10^-4 m = 0.18 mm

22: Problem 22.4

INTRO:
A double-slit experiment is performed with light of wavelength 630nm. The bright interference fringes are spaced 1.6mm apart on the viewing screen.

Part A:
What will the fringe spacing be if the light is changed to a wavelength of 420nm ?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
givens: 
λ_A= 630 nm =
Δy_A = 1.6 mm
λ_B= 420 nm

using the formula Δy = λL⋅1/d we can determine the ratio of the slits' distance from the screen and their distance from each other, or L/d, which would be constant for both wavelengths
using the first wavelength, 
L/d = Δy/λ = Δy_A/λ_A = 1.6 mm / 630 nm = 2.54⋅10^-3 mm/nm
now we can use this to determine the fringe spacing for the second wavelength,
Δy_B = λ_B⋅L/d = (420 nm)⋅(2.54⋅10^-3 mm/nm) 
→Δy_B = 1.067 mm ≈ 1.1 mm