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Section summary

  • Young’s double slit experiment gave definitive proof of the wave character of light.
  • An interference pattern is obtained by the superposition of light from two slits.
  • There is constructive interference when d sin θ = (for m = 0, 1, 1, 2, 2, …) size 12{d`"sin"θ=mλ,`m=0,`1,` - 1,`2,` - 2,` dotslow } {} , where d size 12{d} {} is the distance between the slits, θ size 12{θ} {} is the angle relative to the incident direction, and m size 12{m} {} is the order of the interference.
  • There is destructive interference when d sin θ = m + 1 2 λ (for m = 0, 1, 1, 2, 2, …) size 12{d`"sin"θ= left (m+ { {1} over {2} } right )λ,`m=0,`1,` - 1,`2,` - 2,` dotslow } {} .

Conceptual questions

Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.

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Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.

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Is it possible to create a situation in which there is only destructive interference? Explain.

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[link] shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

The figure shows a photo of a horizontal line of equally spaced red dots of light on a black background. The central dot is the brightest and the dots on either side of center are dimmer. The dot intensity decreases to almost zero after moving six dots to the left or right of center. If you continue to move away from the center, the dot brightness increases slightly, although it does not reach the brightness of the central dot. After moving another six dots, or twelve dots in all, to the left or right of center, there is another nearly invisible dot. If you move even farther from the center, the dot intensity again increases, but it does not reach the level of the previous local maximum. At eighteen dots from the center, there is another nearly invisible dot.
This double slit interference pattern also shows signs of single slit interference. (credit: PASCO)
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Problems&Exercises

At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?

0 . 516º size 12{0 "." "516"°} {}

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Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.

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What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of 30 . size 12{"30" "." 0°} {} ?

1 . 22 × 10 6 m size 12{1 "." "22" times "10" rSup { size 8{ - 6} } `m} {}

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Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of 45 . size 12{"45" "." 0°} {} .

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Calculate the wavelength of light that has its third minimum at an angle of 30 . size 12{"30" "." 0°} {} when falling on double slits separated by 3 . 00 μm size 12{3 "." "00"`"μm"} {} . Explicitly, show how you follow the steps in Problem-Solving Strategies for Wave Optics .

600 nm

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What is the wavelength of light falling on double slits separated by 2 . 00 μm size 12{2 "." "00"`"μm"} {} if the third-order maximum is at an angle of 60 . size 12{"60" "." 0°} {} ?

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At what angle is the fourth-order maximum for the situation in [link] ?

2 . 06º size 12{2 "." "06"°} {}

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What is the highest-order maximum for 400-nm light falling on double slits separated by 25 . 0 μm size 12{"25" "." 0`"μm"} {} ?

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Find the largest wavelength of light falling on double slits separated by 1 . 20 μm size 12{1 "." "20"`"μm"} {} for which there is a first-order maximum. Is this in the visible part of the spectrum?

1200 nm (not visible)

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What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?

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(a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?

(a) 760 nm

(b) 1520 nm

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(a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of 10 . size 12{"10" "." 0°} {} , at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest-order maximum possible here?

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[link] shows a double slit located a distance x size 12{x} {} from a screen, with the distance from the center of the screen given by y size 12{y} {} . When the distance d size 12{d} {} between the slits is relatively large, there will be numerous bright spots, called fringes. Show that, for small angles (where sin θ θ size 12{"sin"θ approx θ} {} , with θ size 12{θ} {} in radians), the distance between fringes is given by Δ y = / d size 12{Δy=xλ/d} {} .

The figure shows a schematic of a double slit experiment. A double slit is at the left and a screen is at the right. The slits are separated by a distance d. From the midpoint between the slits, a horizontal line labeled x extends to the screen. From the same point, a line angled upward at an angle theta above the horizontal also extends to the screen. The distance between where the horizontal line hits the screen and where the angled line hits the screen is marked y, and the distance between adjacent fringes is given by delta y, which equals x times lambda over d.
The distance between adjacent fringes is Δ y = / d size 12{Δy=xλ/d} {} , assuming the slit separation d size 12{d} {} is large compared with λ size 12{λ} {} .

For small angles sin θ tan θ θ ( in radians ) size 12{"sin"θ - "tan"θ approx θ` \( "in"`"radians" \) } {} .

For two adjacent fringes we have,

d sin θ m = size 12{d`"sin"θ rSub { size 8{m} } =mλ} {}

and

d sin θ m + 1 = m + 1 λ size 12{d`"sin "θ rSub { size 8{m+1} } = left (m+1 right )λ} {}

Subtracting these equations gives

d sin θ m + 1 sin θ m = m + 1 m λ d θ m + 1 θ m = λ tan θ m = y m x θ m d y m + 1 x y m x = λ d Δ y x = λ Δ y = d alignl { stack { size 12{d left ("sin"θ rSub { size 8{m+1} } - " sin"θ rSub { size 8{m} } right )= left [ left (m+1 right ) - m right ]λ} {} # d left (θ rSub { size 8{m+1} } - θ"" lSub { size 8{m} } right )=λ {} #"tan"θ rSub { size 8{m} } = { {y rSub { size 8{m} } } over {x} } approx θ"" lSub { size 8{m} } drarrow d left ( { {y rSub { size 8{m+1} } } over {x} } - { {y rSub { size 8{m} } } over {x} } right )=λ {} # d { {Δy} over {x} } =λ drarrow {underline {Δy= { {xλ} over {d} } }} {}} } {}
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Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in [link] .

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Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see [link] ).

450 nm

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Questions & Answers

Why is there no 2nd harmonic in the classical electron orbit?
Shree Reply
how to reform magnet after been demagneted
Inuwa Reply
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
Morris Reply
what is the error during taking work done of a body..
Aliyu Reply
what kind of error do you think? and work is held by which force?
Daniela
I am now in this group
smart
theory,laws,principles and what-a-view are not defined. why? you
Douglas Reply
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
Emmanuel Reply
so what question are you passing across... sir?
Olalekan
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
Emmanuel Reply
54 joule
babar
how?
rakesh
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
rakesh
i dont think dere is any potential energy... by d virtue of no height present
Olalekan
there is compressed energy,dats only potential energy na?
rakesh
yes.. but... how will u approach that question without The Height in the question?
Olalekan
Can you explain how you get 54J?
Emmanuel
Because mine is 36J
Emmanuel
got 36J too
Douglas
OK the answer is 54J Babar is correct
Emmanuel
Conservation of Momentum
Emmanuel
woow i see.. can you give the formula for this
joshua
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel Reply
Please help!
Emmanuel
please help find dy/dx 2x-y/x+y
Inuwa
By using the Quotient Rule dy/dx = 3y/(x +y)²
Emmanuel
3y/(x+y)²
Emmanuel
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
rakesh
i think i m correct
rakesh
But how?
Emmanuel
3y/(x+y)²
Douglas
what's the big bang?
kwame Reply
yes what is it?
LamaBbake
it is the explanation of how the universe began
Zainab
yes
Ana
explain
Chinagorom
in
Chinagorom
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
david
thanks guys
kwame
if a force of 12N is applied to load of 200g what us the work done
Joshua Reply
We can seek accelation first
Nancy
we are given f=12 m=200g which is 0.2kg now from 2nd law of newton a= f/m=60m/s*2 work done=force applied x displacement cos (theta) w= 12x60 =720nm/s*2
Mudang
this very interesting question very complicated for me, í need urgent help. 1,two buses A and B travel along the same road in the same direction from Harper city (asume They both started from the same point) to Monrovia. if bus A maintains a Speedy of 60km/h and bus B a Speedy of 75km/h, how many
mohammed
hours Will it take bus B to overtake bus A assuming bus B starts One hour after bus A started. what is the distance travelled by the buses when They meet?.
mohammed
pls í need help
mohammed
4000 work is done
Ana
speed=distance /time distance=speed/time
Ana
now use this formula
Ana
what's the answer then
Julius
great Mudang
Kossi
please Ana explain 4000 ?
babar
hey mudang there is a product of force and acceleration not force and displacement
babar
@Mohammed answer is 0.8hours or 48mins
Douglas
nice
A.d
its not possible
Olalekan
í want the working procedure
mohammed
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
mohammed
physics is the science that studies the non living nature
isidor Reply
ancient greek language physis = nature
isidor
what is phyacs
technical Reply
if i am going to start studying physics where should i start?
BRIAN Reply
I think from kinematics
Nancy
You can find physics books at the library or online. That's how I started.
Chelsea
And yes, kinematics is usually where you can begin.
Chelsea
study basic algebra and calculus and can start from classical mechanics
Mudang
yes think so but dimension is the best starting point
Obed
3 formula's of equations of motion
benjamin Reply
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
Ana
solve the formula's please
benjamin
those are the three .. what you wanna solve ?
Nihrantz
For first equation simply integrate formula of acceleration in the limit v and u
Tripti
For second itegrate velocity formula by ising first equation
Tripti
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
Tripti
any methods can take to solve this eqtions
a=vf-vi/t vf-vi=at vf=vi+at......1
Ana
suppose a body starts with an initial velocity vi and travels with uniform acceleration a for a period of time t.the distance covered by a body in this time is "s" and its final velocity becomes vf
Ana
what is the question dear
Zeeshan
average velocity=(vi+vf)/2 distance travelled=average velocity ×time therefore s=vi+vf/2×t from the first equation of motion ,we have vf =vi+at s=[vi+(vi+at)]/2×t s=(2vi+at)/2×t s=bit+1/2at2
Ana
find the distance
Ana
how
Zeeshan
Two speakers are arranged so that sound waves with the same frequency are produced and radiated through a room. An interference pattern is created. Calculate the distance between the two speakers?
Hayne Reply
How can we calculate without any information?
Amir
I think the formulae used for this question is lambda=(ax)/D
Amir
Practice Key Terms 5

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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