1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Intensity from the double slit.

  1. Oct 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Laser light of wavelength λ is incident onto a pair of narrow slits of width a and spacing d

    Write down an expression for the intensity as a function of angle of the angle measured from the central axis, θ, in the far-*‐field

    2. Relevant equations

    3. The attempt at a solution

    I thought it was just the product of single and double slit diffraction, i.e....

    I= I0cos2(∏dsinθ/λ)(sin(∏asinθ/λ)/∏asinθ/λ))2

    but the answer was I= 4I0cos2(∏dsinθ/λ)(sin(∏asinθ/λ)/∏asinθ/λ))2

    im not sure where this 4 has come from??

    also how would I show if d=a then this formula turns into the double slit pattern.??
    Last edited: Oct 29, 2012
  2. jcsd
  3. Oct 29, 2012 #2


    User Avatar
    2017 Award

    Staff: Mentor

    The prefactor depends on the definition of I0.
    If you define it as "maximal intensity with a double-slit", the prefactor is 1.
    If you define it as "maximal intensity with a single slit", it is 4.

    If the distance between the slit centers is equal to the slit size, why would you expect a double slit pattern?
  4. Oct 29, 2012 #3

    sorry it should have been single slit with width 2a... but when I plug in d=a I can't see how that could disappear?
  5. Oct 30, 2012 #4


    User Avatar
    2017 Award

    Staff: Mentor

    With ##x=\frac{\pi a \sin(\theta)}{\lambda}##:

    $$I= I_0\cos^2\left(\frac{\pi d \sin(\theta)}{\lambda}\right) \frac{sin^2(x)}{x^2}$$
    With d=a and using sin(2x)=2sin(x)cos(x):
    $$I= I_0 \frac{\cos^2(x) sin^2(x)}{x^2} = I_0 \frac{1}{4} \frac{sin^2(2x)}{x^2} = I_0 \frac{sin^2(2x)}{(2x)^2}$$
    Which is the single-slit pattern with 2a.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook