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.

 

[Nickie]

I would like to clarify that the expression provided in the attempt at a solution is not entirely correct. The correct expression for the intensity from a double slit is given by the equation:

I = I0(cos(πdsinθ/λ)/πdsinθ/λ)^2

Where I0 is the intensity at the central maximum, d is the slit spacing, λ is the wavelength, and θ is the angle measured from the central axis.

Now, to address the question about where the factor of 4 comes from, it is a result of the double slit interference pattern. The intensity at a point on the screen is given by the sum of the intensities from each slit. Since there are two slits, the intensity is doubled, resulting in the factor of 4.

To show that when d=a, the expression turns into the double slit pattern, we can substitute d=a into the equation and simplify it. Doing so will result in the expression:

I = 4I0(cos(πasinθ/λ)/πasinθ/λ)^2

This is the same expression as the one for a single slit, but with a factor of 4 multiplied to it. This indicates that when the slit spacing is equal to the slit width, the double slit pattern reduces to the single slit pattern with a higher intensity.

In conclusion, the correct expression for the intensity from a double slit is I = I0(cos(πdsinθ/λ)/πdsinθ/λ)^2 and the factor of 4 in this equation is a result of the double slit interference pattern. When d=a, the expression reduces to the single slit pattern with a higher intensity.