Single slit with minimum uncertainty

[HelpPlease27]

Homework Statement

A horizontal beam of laser light of wavelength 474 nm passes through a narrow slit that has width 5.80×10−2 mm . The intensity of the light is measured on a vertical screen that is 2.40 m from the slit.
Use the result of part A to estimate the width of the central diffraction maximum that is observed on the screen.
(Part A: What is the minimum uncertainty in the vertical component of the momentum of each photon in the beam after the photon has passed through the slit? I got this part.)

Homework Equations

$$\theta = \arcsin(\lambda/a)$$
$$y = d * \tan\theta$$

The Attempt at a Solution

I just used the two equations above first to get theta and then to get y and times y by two to get the central diffraction maximum but that's wrong. I'm missing at which part I'm supposed to use the minimum uncertainty. Can anyone explain, please?

 

[TSny]

It could be that you are expected to answer part (B) without using the optics single slit diffraction formulas. Instead, use the result of part (A). Treat the photon as traveling in the x direction before reaching the slit. As it passes through the slit, it picks up some y-component of momentum that you can estimate from your answer in part (A). (Careful with factors of 2, etc.) Use the x and y components of momentum of the photon as it leaves the slit to estimate the direction the photon heads out from the slit. Therefore, estimate where it will hit the screen.

 

[HelpPlease27]

I don't understand how I can use that to work out the central diffraction maximum. $$\Delta Py*\Delta y<h/4\pi$$ So is $$\Delta y$$ the central diffraction maximum? Where does the x component come into it?

 

[TSny]

I don't understand how I can use that to work out the central diffraction maximum. $$\Delta Py*\Delta y<h/4\pi$$ So is $$\Delta y$$ the central diffraction maximum? Where does the x component come into it?

$\Delta y$ is the uncertainty in the y position of the photon as it passes through the slit. You must have used this idea when working part (A).

$\Delta P_y$ is the corresponding uncertainty in y-component of momentum that the photon acquires in order to satisfy the Heisenberg uncertainty principle. It's your answer to (A).

You are only doing an estimate. So, roughly speaking, you can say that the photons leave the slit with a y-component of momentum that could be anything between $+\Delta P_y/2$ and $- \Delta P_y/2$. So, photons will arrive at the screen with a spread in the y direction which we assume corresponds roughly with the central maximum.

 

[HelpPlease27]

You are only doing an estimate. So, roughly speaking, you can say that the photons leave the slit with a y-component of momentum that could be anything between $+\Delta P_y/2$ and $- \Delta P_y/2$.

I get this part but not the next part.

So, photons will arrive at the screen with a spread in the y direction which we assume corresponds roughly with the central maximum.

I still don't get what the spread in the y-direction corresponds to?

 

[TSny]

The direction the photon travels from the slit to the screen is the direction of the momentum vector of the photon. This direction is determined by the x and y components of the momentum: $P_x$ and $P_y$.

 

[HelpPlease27]

Thanks for explaining it

 

[TSny]

OK. I think it would have been better if I had said that you can determine the direction θ from the magnitude of the momentum of the photon and the y-component of the momentum.

 

[HelpPlease27]

So then is $$\tan\theta * x = y$$ which is the central diffraction maximum? Or am I still missing something?

 

[TSny]

So then is $$\tan\theta * x = y$$ which is the central diffraction maximum? Or am I still missing something?

This equation will give a rough estimate of the distance on the screen from the center of the central max to the first minimum. The width of the central maximum will be twice this distance.

 

[HelpPlease27]

Ok, I get it. Thank you