Jonsson’s apparatus for photons rather than electrons

[Phys12]

TL;DR

What values would d,D,and w take if Jonsson’s apparatus were simply scaled up for use with visible light rather than electrons?

In the prompt 4c in this problem set: https://ocw.mit.edu/courses/physics...-i-spring-2013/assignments/MIT8_04S13_ps1.pdf. We are asked to find out how d, D and w will change (where d is the slit width, D is the distance from the slits to the screen and w is the distance between adjacent maxima in the interference pattern) if the apparatus were to be scaled up for use with visible light rather than electrons. In part 4a, we found out that the relationship between these variables is $w = \frac{Dh}{dp}$ or $w = \frac{D \lambda}{dp}$. The answer for 4c is (https://ocw.mit.edu/courses/physics...pring-2013/assignments/MIT8_04S13_ps1_sol.pdf) that we would scale each quantity by the ratio of the wavelength of the photon to the wavelength of the electron used earlier. Why is that the case? I don't think I understand what it means to scale up the apparatus for use with visible light. Why would you want to scale it up? To make sure that w stays the same? But if that were the case, in the answer, w's value would be the same...

 

[BvU]

Why would you want to scale it up?

It's a 'what if ' question to show that the electron experiment is quite sophisticated.

in the answer, w's value would be the same...

No: $$w = \lambda \,{D\over d}$$so $w$ scales with $\lambda$.

 

[Phys12]

It's a 'what if ' question to show that the electron experiment is quite sophisticated.

But why do you need to change D, w, and d in order to incorporate photons? I thought the only thing that needed to change was the slit width to actually make the diffraction happen...

No: $$w = \lambda \,{D\over d}$$so $w$ scales with $\lambda$.

Right, but if you scale $d$ and $D$ by the same amount, wouldn't the two $a$'s cancel to give you the original $w$?

 

[BvU]

But why do you need to change D, w, and d in order to incorporate photons? I thought the only thing that needed to change was the slit width to actually make the diffraction happen...

The point they are trying to make is that the angle is really small. There is no one who would actually put the screen at 35 km...

Right, but if you scale $d$ and $D$ by the same amount, wouldn't the two $a$'s cancel to give you the original $w$?

those two $a$ cancel yes, but there still is a scale factor from $\lambda$ itself

All they are really trying to bring across here is that wavelengths for even the lightest particles are very, very small.