Question

A beam of particles, each of mass m and (nonrelativistic) speed v, strikes a barrier in which there are two narrow slits and beyond which is a bunk of detectors. With slit 1 alone open, 100 particles are detected per second at all detectors. Now slit 2 is also opened. An interference pattern is noted in which the first minimum. 36 particles per second. Occurs at an angle of 30^ofrom the initial direction of motion of the beam.

(a) How far apart are the slits?

(b) How many particles would be detected ( at all detectors) per second with slit 2 alone open?

(c) There are multiple answers to part (b). For each, how many particles would be detected at the center detector with both slits open?

Short answer

(a)In terms of the provided quantities, slit separation is$\text{d=}\frac{\text{h}}{\text{mv}}$

(b)The number of electrons per second in a particular area for slit 2 can be either 16 electrons per second or 256 electrons per second.

(c)With both slits open, a signal was caught at the central detectoris either 196 or 676 electrons per second.

Step-by-step solution

Concept Introduction

The special difference between constructive and destructive interference is that when the displacements of the waves that meet are in the same direction, whereas destructive interference occurs when displacements of the waves that meet are in the opposite directions.

Find Slit Separation.

(a)

We may find the slit separation in terms of the provided quantities by using the double-slit interference relation (m and v)

$\text{dsin(θ)=}\left(\text{n+}\frac{\text{1}}{\text{2}}\right)\text{λ}$

$\text{λ=}\frac{\text{h}}{\text{p}}$

$\text{d=}\frac{\left(\text{n+}\frac{\text{1}}{\text{2}}\right)\text{h}}{\text{sin(θ)p}}$

$\text{n=0,\hspace{0.17em}sin(30)=}\frac{\text{1}}{\text{2}}\text{,\hspace{0.17em}p=mv}$

$\text{d=}\frac{\text{h}}{\text{mv}}$

Calculation of the counts for the slit 2

Let's look at the data in the problem: 100 electrons per second are observed with slit one alone open (the corresponding wave is.${\text{ψ}}_{\text{1}})$However, with slit two open, 36 electrons per second are detected at the first minimum (the corresponding wave is${\text{ψ}}_{\text{2}})$. When the two waves destructively interfere with each other, this minimum occurs$\left({\text{ψ}}_{\text{1}}{\text{-ψ}}_{\text{2}}\right)$

$\begin{array}{l}{\left|{\text{ψ}}_{\text{1}}\right|}^{\text{2}}\propto \text{100}\\ \left|{\text{ψ}}_{\text{1}}\right|\propto \text{10}\end{array}$

and

$\begin{array}{l}{\left|{\text{ψ}}_D\right|}^{\text{2}}\propto 36\\ \left|{\text{ψ}}_D\right|\propto \pm 6\end{array}$

For destructive interference, we can write,

$\begin{array}{l}\left|{\text{ψ}}_{\text{D}}\right|\propto \left|{\text{ψ}}_1\right|-\left|{\text{ψ}}_2\right|\\ \left|{\text{ψ}}_2\right|\propto \left|{\text{ψ}}_{\text{D}}\right|+\left|{\text{ψ}}_1\right|\\ \left|{\text{ψ}}_{\text{2}}\right|\propto 10\pm 6\end{array}$

$\begin{array}{c}{\left|{{\text{ψ}}_{\text{2}}}^{\text{'}}\right|}^2\propto {\left(16\right)}^2\\ \propto 256\end{array}$

$\begin{array}{c}{\left|{{\text{ψ}}_{\text{2}}}^{\text{'}\text{}\text{'}}\right|}^2\propto {\left(4\right)}^2\\ \propto 16\end{array}$

The number of electrons per second in a particular area for slit 2 can be either 16 electrons per second or 256 electrons per second.

Center Detector

The constructive interference means the addition of the two-slit wave functions together. As we already have the amplitude of each wave we can directly find the total number of electrons detected per second in a given area for each case of part b).

Therefore, considering both possibilities discussed in part (b), we can write,

For$\left|{{\text{ψ}}_{\text{2}}}^{\text{'}}\right|$

$\begin{array}{c}{\left|{\text{ψ}}_{\text{1}}{{\text{+ψ}}_{\text{2}}}^{\text{'}}\right|}^{\text{2}}\propto {\left(\text{10+16}\right)}^{\text{2}}\\ \propto {\left(\text{26}\right)}^{\text{2}}\\ \propto \text{676}\end{array}$

For

$\begin{array}{c}{\left|{\text{ψ}}_{\text{1}}{{\text{+ψ}}_{\text{2}}}^{\text{'}\text{}\text{'}}\right|}^{\text{2}}\propto {\left(\text{10+4}\right)}^{\text{2}}\\ \propto {\left(\text{14}\right)}^{\text{2}}\\ \propto \text{256}\end{array}$

Results

(a)In terms of the provided quantities, slit separation is$\text{d=}\frac{\text{h}}{\text{mv}}$

(b) The number of electrons per second in a particular area for slit 2 can be either 16 electrons per second or 256 electrons per second.

(c) With both slits open, a signal was caught at the central detector is either 196 or 676 electrons per second.