High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Physics Textbook (Master) > Wave Optics

Double Slit Diffraction

Light Waves at BoundariesPrevious Section
Double Slit Diffraction Modified Next Section
Popular Courses
Find My Course

Double-Slit Interference


The principle of double-slit interference is similar to the interference of sound waves from two speakers.




PAGE BREAK

Bright fringes on the screen correspond to constructive interference. The path difference Δl\Delta l is given by:
 dsin⁡θ=nλ \boxed{ \ d\sin\theta=n\lambda \ }
n=0,±1,±2,…n=0,\pm1,\pm2,\ldots
  • dd is the distance between the two slits
  • θ\theta is the angle that the rays make with the horizontal

Wize Concept
The formula above is valid only when d≪Dd\ll D.






Dark fringes on the screen correspond to destructive interference. The path difference Δl\Delta l is given by:
 dsin⁡θ=(n+12)λ \boxed{ \ d\sin\theta=\bigg(n+\dfrac{1}{2}\bigg)\lambda \ }
n=0,±1,±2,…n=0,\pm1,\pm2,\ldots

An equivalent way of writing this would be:
 dsin⁡θ=n′ λ2 \boxed{ \ d\sin\theta=n' \ \dfrac{\lambda}{2} \ }
n′=0,±3,±5,…    (odd)n'=0,\pm3,\pm5,\ldots \ \ \ \ (odd)




Exam Tip
The zero order maximum occurs in the middle, so you begin labeling on either side of it the alternating bright and dark fringes, as nn increases.




PAGE BREAK

Let's define the following quantities:
  • yy is the distance on the screen between the point of interference and the central bright spot
  • DD is the distance between the slits and the screen
Then we have a triangle in which:
 tan⁡θ=yD \boxed{ \ \tan\theta=\frac{y}{D} \ }

Wize Tip

When y≪Dy \ll D we can use the small angle approximation:
 θ≈sin⁡θ≈tan⁡θ \boxed{ \ \theta\approx\sin\theta\approx\tan\theta \ }



Example: Number of Fringes


A double-slit configuration is shown below, with d=1d=1 mm. Light with a wavelength of 400400 nm is used.

a) How many (bright) fringes are displayed on the screen?
b) How many (bright) fringes would be displayed on the screen if we double its length (L=2L=2m)?
c) What if the length of the screen is unlimited?


Part a)


For bright fringes we have constructive interference, so we use:

dsin⁡θ=mλd\sin\theta=m\lambda

Since y<<Dy<<D we can use the small angle approximation: yD=tan⁡θ≈sin⁡θ\dfrac{y}{D}=\tan\theta\approx\sin\theta to get:

d⋅yD=mλd\cdot\dfrac{y}{D}=m\lambda

The maximum distance on the screen is half the length of the screen, so we'll use y=0.5y=0.5 to find the maximum that mm can be:

m=ydλD=(0.5)(0.001)(10)(400×10−9)=125m=\dfrac{yd}{\lambda D}=\dfrac{(0.5)(0.001)}{(10)(400\times10^{-9})}=125

But this means we have 125125 fringes on either side, and we also have to add the central bright fringe (corresponding to m=0m=0), so we have a total of:

125⋅2+1=251125\cdot2+1=251 bright fringes


Part b)


If the length of the screen doubles, the yy doubles, and the mm we solved for also doubles (because they are directly proportional in the equation), so we get m=250m=250.

Again, we have 250250 bright fringes on either side, and if we also count the central bright fringe, the total is:

250⋅2+1=501250\cdot2+1=501 bright fringes


Part c)


When the length of the screen is infinite, we could have a maximum angle of 90°\bcth{90\degree}, and sin⁡90°=1\sin90\degree=1. Therefore our equation becomes:

dsin⁡θ=mλd\sin\theta=m\lambda

d(1)=mmaxλd(1)=m_{max}\lambda

And we get:

mmax=dλ=0.001400×10−9=2500m_{max}=\dfrac{d}{\lambda}=\dfrac{0.001}{400\times10^{-9}}=2500


We have 25002500 bright fringes on either side, so the total is:

2500⋅2+1=50012500\cdot2+1=5001 bright fringes

Practice: Argon Laser

A two-slit diffraction grating, with slits at distance of 0.20.2 mm apart is illuminated by an Argon laser (wavelength 454.6454.6 nm).

a) What is the angle between the two second brightest points on a screen placed 2020 cm away from the slits? (i.e. between the first bright spot immediately above the central max, and the first bright spot immediately below the central max)
b) What is the corresponding distance between these two points?
c) What is the distance between the third and fourth brightest points on the screen?

Double-Slit Interference

Bright fringes on the screen correspond to constructive interference.
 dsin⁡θ=nλ \boxed{ \ d\sin\theta=n\lambda \ }
n=0,±1,±2,…n=0,\pm1,\pm2,\ldots
  • dd is the distance between the two slits
  • θ\theta is the angle that the rays make with the horizontal

Dark fringes on the screen correspond to destructive interference. The path difference Δl\Delta l is given by:
 dsin⁡θ=(n+12)λ \boxed{ \ d\sin\theta=\bigg(n+\dfrac{1}{2}\bigg)\lambda \ }
n=0,±1,±2,…n=0,\pm1,\pm2,\ldots

Exam Tip
The zero order maximum occurs in the middle, so you begin labeling on either side of it the alternating bright and dark fringes, as nn increases.


Let's define the following quantities:
  • yy is the distance on the screen between the point of interference and the central bright spot
  • DD is the distance between the slits and the screen
Then we have a triangle in which:
 tan⁡θ=yD \boxed{ \ \tan\theta=\frac{y}{D} \ }

Wize Tip

When y≪Dy \ll D we can use the small angle approximation:
 θ≈sin⁡θ≈tan⁡θ \boxed{ \ \theta\approx\sin\theta\approx\tan\theta \ }





A diffraction grating has multiple slits located the same distance apart. But same principle and same formulas work as of the double-slit.


If you're given number of lines per unit length (NN) , the distance between each two slits could be found from:
 d=1N \boxed{ \ d=\frac{1}{N} \ }


Exam Tip
Just remember that NN and dd are reciprocals of each other.


Watch Out!
Often you're given the number of lines per mmmm or cmcm. This means the reciprocal will give you dd in those units as well, so you'll have to convert it to meters.