Recalling the Equation for Slit Width in a Double Slit Interference Experiment

  • Thread starter Thread starter KBriggs
  • Start date Start date
AI Thread Summary
The discussion centers on finding the equation for the slit width in a double slit interference experiment, specifically in relation to the angle theta of the interference maxima. A participant suggests that the equation for fringe width, W = (Dλ)/S, may be relevant, where W is fringe width, D is the distance from the slits to the screen, λ is the wavelength of light, and S is the slit separation. However, the original poster clarifies that they are seeking the actual size of the slits rather than fringe width. The conversation highlights the need for a formula that directly relates to slit width without relying on intensity measurements. The request for assistance remains open as the specific equation for slit width is not provided.
KBriggs
Messages
29
Reaction score
0
Hey all,

I need to get the equation for the slit width in a double slit interference experiment (not the distance between slits) that depends on the angle theta to the intferference maxima, and any distances involved in the typical setup. I found one on the forum that involves the intensity of the maxima, but I have no way to measure intensity. Any help?
 
Physics news on Phys.org
KBriggs said:
Hey all,

I need to get the equation for the slit width in a double slit interference experiment (not the distance between slits) that depends on the angle theta to the intferference maxima, and any distances involved in the typical setup. I found one on the forum that involves the intensity of the maxima, but I have no way to measure intensity. Any help?

well do you mean fringe width over here..!if yes then equation goes like this..

W = (Dλ)/S

Where,
W= fringe width
D=distance from slits to screen
λ= WAVELENGTH of light
S=slit seperation(distance between the slits)..

I hope this is the formula you were looking for.
 
Not the fringe width, but the size of the two slits. ie, the size of the larger gaps in the little diagram here.
|
|
|
|

|
|
|

|
|
|
|
 
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (First part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8 and stuck at some statements. It's little bit confused. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. Solution : The surface bound charge on the ##xy## plane is of opposite sign to ##q##, so the force will be...
Back
Top