Optics: Intensity Homework - Solving for Distance to Io/2

[Ginerva123]

Homework Statement

Parallel rays of light pass of wavelength 540nm throught a 0.4mm slit. The intensity of the central maximum on a diffraction pattern 1.5m away is Io. At what distance from the central maximum does the intensity fall to Io/2?

Homework Equations

I = Io sin^2 (pi a y / lambda d) / (pi a y / lambda d)^2 (?)

The Attempt at a Solution

Okay, I worked out the distance from the central maximum to the first minimum to be
2.03mm, so it's obviously less than that... beyond that, I'm stuck. Any ideas?

 

[anathema]

Thank you for your question. Based on your calculations, it seems that you have already determined the distance from the central maximum to the first minimum, which is 2.03mm. To find the distance from the central maximum to where the intensity falls to Io/2, we can use the equation for intensity, which you have correctly listed as I = Io sin^2 (pi a y / lambda d) / (pi a y / lambda d)^2.

To find the distance, we can rearrange this equation to solve for y. This gives us y = lambda d / (pi a) * sqrt(-ln(I/Io)), where I is the desired intensity (in this case, Io/2). Plugging in the values for lambda, d, and a, we get y = 1.08mm as the distance from the central maximum to where the intensity falls to Io/2.

I hope this helps! Let me know if you have any further questions.

 

[Moetasim]

To solve for the distance where the intensity falls to Io/2, we can use the formula for intensity in the diffraction pattern as follows:

I = Io sin^2 (pi a y / lambda d) / (pi a y / lambda d)^2

Where:
- I is the intensity at a given point in the diffraction pattern
- Io is the intensity of the central maximum
- a is the slit width
- y is the distance from the central maximum
- lambda is the wavelength of light
- d is the distance from the slit to the diffraction pattern

To solve for the distance where the intensity falls to Io/2, we can rearrange the equation to solve for y:

y = (lambda d / pi a) * sqrt(ln(2))

Plugging in the given values, we get:

y = (540nm * 1.5m / (3.14 * 0.4mm)) * sqrt(ln(2)) = 0.0018m = 1.8mm

Therefore, the distance where the intensity falls to Io/2 is 1.8mm from the central maximum.