Intensity from the double slit.

[sg001]

Homework Statement

Laser light of wavelength λ is incident onto a pair of narrow slits of width a and spacing dWrite down an expression for the intensity as a function of angle of the angle measured from the central axis, θ, in the far-*‐field

Homework Equations

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

 

[mfb]

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?

 

[sg001]

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?

 

[mfb]

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.