Young's Double Slit Experiment

In summary, the conversation discusses the equation for the interference pattern created by a single slit. The variable m represents the number of spots away from the center, with m being an integer according to both the teacher and other sources. The equation simplifies to (\frac{1}{2})\gamma =dsin(\theta), (\frac{3}{2})\gamma =dsin(\theta), (\frac{5}{2})\gamma =dsin(\theta)... for m = 0, 1, 2... respectively. This fact is further clarified by discussing the first, second, and third spots and their corresponding equations.
  • #1
nesan
75
0

Homework Statement


Is this wrong or right? PLZ.

The Attempt at a Solution



http://www.kentshillphysics.net/optics36.gif

I have a test tomorrow buy my teacher said it's m - 1/2

Why does it say n + 1/2 everywhere. >_<
 
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  • #2
What is m in this equation? (both according to your teacher and to the other sources you found)
 
  • #3
Villyer said:
What is m in this equation? (both according to your teacher and to the other sources you found)

According to my teacher n is the number of spots from the center (dark or light).

Here's one of my other source.

http://farside.ph.utexas.edu/teaching/316/lectures/node151.html
 
  • #4
From your source, m is an integer 0,1,2...
This means that the equation simplyfies into [itex](\frac{1}{2})\gamma =dsin(\theta)[/itex], [itex](\frac{3}{2})\gamma =dsin(\theta)[/itex], [itex](\frac{5}{2})\gamma =dsin(\theta)[/itex]...

From your teacher, the number of spots away from the center is n.

So the first spot would yield [itex](\frac{1}{2})\gamma =dsin(\theta)[/itex], the second would be [itex](\frac{3}{2})\gamma =dsin(\theta)[/itex], and the third would be [itex](\frac{5}{2})\gamma =dsin(\theta)[/itex].

Do you see where I'm going with this?
 
  • #5
Villyer said:
From your source, m is an integer 0,1,2...
This means that the equation simplyfies into [itex](\frac{1}{2})\gamma =dsin(\theta)[/itex], [itex](\frac{3}{2})\gamma =dsin(\theta)[/itex], [itex](\frac{5}{2})\gamma =dsin(\theta)[/itex]...

From your teacher, the number of spots away from the center is n.

So the first spot would yield [itex](\frac{1}{2})\gamma =dsin(\theta)[/itex], the second would be [itex](\frac{3}{2})\gamma =dsin(\theta)[/itex], and the third would be [itex](\frac{5}{2})\gamma =dsin(\theta)[/itex].

Do you see where I'm going with this?

Oh I got it. :)

Tnx. :)

<3
 

Related to Young's Double Slit Experiment

1. What is Young's Double Slit Experiment?

Young's Double Slit Experiment is a classic physics experiment that demonstrates the wave-like nature of light. It involves shining a beam of light through two narrow slits and observing the resulting interference pattern on a screen.

2. What does the experiment demonstrate?

The experiment demonstrates the wave-like behavior of light, as the interference pattern observed on the screen can only be explained by the superposition of light waves from the two slits.

3. How does the interference pattern form?

The interference pattern is formed when the waves from the two slits overlap and either reinforce or cancel each other out, creating bright and dark fringes on the screen.

4. What factors affect the interference pattern?

The interference pattern is affected by the wavelength of light, the distance between the slits, and the distance between the slits and the screen.

5. What can the experiment tell us about the nature of light?

The experiment provides evidence for the wave-like nature of light and supports the theory of quantum mechanics, which describes light as both a wave and a particle.

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