More Light Questions
1.
You are performing Young’s Double-Slit Experiment. State three ways to increase
the distance between bright fringes that appear on the screen.
Move screen farther away, make slits closer together, use longer wavelength light (closer
to the red end of spectrum)
2.
Light travels at a speed of 1.95 x 108 m/s in a certain medium. What is the
wavelength of green light in this medium if it has a wavelength of 560 nm in air?
n = c/v so
n = 3.00/1.95 = 1.54
λmed = λair / n = 560/1.54 ... λmed = 363nm
Remember that any wave will be shorter wavelength in a new medium compared to air,
since it will always slow down in that new medium. When waves are slowed down, they
are compressed = shortened wavelength.
3.
What is the wavelength of a microwave if it has a frequency of 2780 MHz?
v=fλ
λ = v /f = 3.0 x 108 / 2780 x 106 = 0.108m
about 11cm.
4.
You are standing in a room facing two speakers that are 3m apart. You are
standing
4m from one speaker and 6.34m from the other. The speakers are
sending a 440Hz
signal. Is the sound interfering constructively or destructively at
the place you are standing?
(The speed of sound in air is 343m/s.)
v = f λ so one wavelength λ = v/f = 343 / 440 = 0.78m
One speaker is farther from you than the other: 6.34m – 4.0m = 2.34m
This is the path difference.
2.34m is equal to exactly 3 wavelengths (3 x 0.78m)
Sound arrives in phase. You will hear constructive interference.
Note: the fact that the speakers are 3m apart has nothing to do with the solution.
5.
In a ripple tank, a point located on the third nodal line from the centre line is
45.0 cm from one source and 60.0 cm from the other source. The sources vibrate
in phase at 6.0 Hz. Calculate the wavelength of the waves.
A nodal line is where the waves destruct. Since this is the third nodal line:
1st node: path diff = λ/2
2nd node: path diff = 3λ/2
3rd node: path diff = 5λ/2
The path difference is 15cm. So: 15cm = 5λ/2
The waves are 6cm long.
6.
λ = 6cm
A student performing Young's Experiment measures a distance of 3.0 cm between
the first and fifth nodal line on a screen that is located 2.3 m from the slits. If the
separation of the slits is 0.20 mm, then what is the wavelength of the laser light in
nanometres?
1st and 5th nodal line = 4 spaces... n=4
nλ = dx/L
4λ = (0.20x10-3)(3.0x10-2)/2.3
λ = 6.5x10-7
λ = 650 nm
7.
Blue light ( = 482 nm) is directed through two parallel slits separated by 0.12 mm.
A fringe pattern appears on a screen 1.80 m away. How far from the central axis
on either side are the second-order dark bands?
2nd dark band is between 1st and 2nd bright band (Zero = central axis)
nλ = dx/L
1.5(482x10-9) = (0.12x10-3)( x )/1.8
x = 0.0108m
the second dark bands are 1.1cm away from the central axis
8.
Helium-neon laser light ( = 633 nm) passes through a single slit with a width of
4.50 x 10-5 m onto a screen 3.00 m away. What is the width of the central maximum in (i)
centimetres and (ii) degrees?
nλ = wx/L
1(633x10-9) = (4.50x10-5)( x ) / 3.00
x = 0.042m
x = 4.2cm to first dark node. So central max is 8.4cm wide
nλ = wsinθ
1(633x10-9) = (4.50x10-5) sinθ
θ = 0.81o to first dark node. Central max is 1.6o wide
9.
A thin layer of oil (n = 1.25) floats on water (n = 1.33). What is the thickness of
the region that strongly reflects violet light (wavelength of 420 nm) in
air?
* draw a diagram
* both waves go out of phase 180o upon reflection
* we want 420nm light to constructively interfere
* path difference has to be 360o (i.e. λ)
2t = λoil
t = λoil/2
λoil = λair / 1.25 = 420/1.25 = 336 nm
so t = 336nm/2
thickness is 168nm
10.
Explain why Young’s experiment is significant to our understanding of light.
Showed that light is a wave since:
Showed diffraction (spreading of light after going through an opening)
Showed that different colours have different wavelengths
Showed that light interferes with itself.
the oil in