Fraunhofer diffraction, multiple slits

[zimo]

Homework Statement

A system consisting of 5-slit (completely identical) that the distance between their centres, d, is four times the width of a slit, b.

Calculate the relation between the major maximum at the center and the next maximum (can be given that position of the maximum is between the two minima)

Homework Equations

I = E0^2(sin^2(beta)/beta^2)*(sin^2(5gamma)/sin^2(gamma))

The Attempt at a Solution

Tried some stuff with finding out the two minima and receiving 3(lambda)/2b but then when placing it inside the equation I couldn't continue.

 

[anathema]

Thank you for your question. The relation between the major maximum at the center and the next maximum in a system of 5 slits can be calculated using the following formula:

I = I0sin^2(theta)/theta^2 * sin^2(Ngamma)/sin^2(gamma)

Where:
I = intensity of the light at a given point
I0 = maximum intensity at the center of the pattern
theta = angle between the center of the pattern and the point of measurement
N = number of slits
gamma = phase difference between adjacent slits

In this case, N = 5 and gamma = 2pi*d/lambda = 8pi*b/lambda, where d is the distance between the centers of the slits and lambda is the wavelength of the light.

To find the relation between the major maximum at the center and the next maximum, we can set theta = pi/2 (since the maximum is between two minima) and N = 2, as we are interested in the next maximum. This gives us the following equation:

I = I0 * sin^2(5gamma)/sin^2(gamma)

To simplify this equation, we can use the trigonometric identity sin^2x = (1-cos2x)/2. Plugging this in, we get:

I = I0 * (1-cos10gamma)/(1-cos2gamma)

Next, we can use the double angle formula cos2x = 1-2sin^2x to further simplify the equation:

I = I0 * (1-(1-2sin^2(5gamma)))/(1-(1-2sin^2(gamma)))

Simplifying further, we get:

I = I0 * (2sin^2(gamma)-2sin^2(5gamma))/(2sin^2(gamma)-2sin^2(gamma))

Finally, using the trigonometric identity sin^2x = (1-cos2x)/2 again, we get:

I = I0 * (2(1-cos2gamma)-2(1-cos10gamma))/(2(1-cos2gamma)-2(1-cos2gamma))

Simplifying, we get:

I = I0 * (cos2gamma-cos10gamma)/(cos2gamma-cos2gamma)

And finally, using the double angle formula cos2x