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Intensity from the double slit.

  • Thread starter sg001
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1. Homework Statement

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. Homework 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:
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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.

also how would I show if d=a then this formula turns into the double slit pattern.??
If the distance between the slit centers is equal to the slit size, why would you expect a double slit pattern?
 
134
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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?
Thanks,

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?
 
33,493
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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.
 

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