Young's Experiment - (problem with equation)

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Young's double slit experiment when the expression d \sin \theta = m \lambda is satisfied, where d is the distance between the slits, theta is the angle from the central axis, lambda is the wavelength, and m is the index of where the maximum occurs. However, when m = 0, the expression for wavelength becomes invalid because the first maximum always occurs at 0 angle regardless of the wavelength. Therefore, the wavelength cannot be determined from the position of the central maximum.
  • #1
FrogPad
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So my book has the following expression for Young's double slit experiment.

(maxima - bright fringes) [itex] d \sin \theta = m \lambda [/itex]
for [itex] m = 0, \, 1 \, 2 \, \ldots [/itex].So what if you solve this for wavelength.

[tex] \lambda = \frac{d \sin \theta}{m} [/tex]

How is this valid when [itex] m = 0 [/itex]

Is this because if [itex] m = 0 [/itex], [itex] \theta [/itex] HAS to equal 0?

by the way
d is the distance between the slits
theta is the angle from the central axis
lambda is the wavelength
m is the index of where the maxima occur
 
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  • #2
FrogPad said:
So my book has the following expression for Young's double slit experiment.

(maxima - bright fringes) [itex] d \sin \theta = m \lambda [/itex]
for [itex] m = 0, \, 1 \, 2 \, \ldots [/itex].So what if you solve this for wavelength.

[tex] \lambda = \frac{d \sin \theta}{m} [/tex]

How is this valid when [itex] m = 0 [/itex]

Is this because if [itex] m = 0 [/itex], [itex] \theta [/itex] HAS to equal 0?
The first maximum is always at 0 angle regardless of wavelength. So you can't determine the wavelength from the position of the central maximum.

AM
 
  • #3
I understand your concern about the validity of the equation when m=0. However, I would like to clarify that the equation you have provided is a simplified version of the more general equation for Young's double slit experiment, which is given by:

d \sin \theta = n \lambda

where n is the order of the bright fringe.

In this more general equation, n can take on any positive or negative integer value, including 0. This means that when m=0, we are simply considering the central bright fringe, where n=0. In this case, the equation becomes:

d \sin \theta = 0

which is valid when \theta = 0 or \pi. This is because the central bright fringe occurs at the central axis (i.e. \theta = 0) and also at the opposite side of the central axis (i.e. \theta = \pi).

Therefore, the equation you have provided is still valid for m=0, as it is just a simplified version of the more general equation. It is important to keep in mind that equations are often simplified for easier understanding and application, but it is always important to consider the full context and assumptions of the equation.

I hope this explanation has cleared up any confusion you may have had about the validity of the equation. Young's double slit experiment is a fundamental experiment in optics and has been extensively studied and validated, so you can trust in the equations and their applications. As always, keep questioning and exploring, as that is the essence of science!
 

1. What is Young's Experiment and what does it demonstrate?

Young's Experiment, also known as the double-slit experiment, is a classic experiment in optics that demonstrates the wave-like nature of light. It involves a beam of light passing through two slits and creating an interference pattern on a screen, which shows the wave-like behavior of light.

2. What is the problem with the equation used in Young's Experiment?

The problem with the equation used in Young's Experiment is that it does not accurately predict the observed interference pattern when a single photon is used. This is because the equation assumes that light behaves as a wave, but when a single photon is used, it behaves as a particle, leading to a different pattern.

3. How does Young's Experiment relate to the wave-particle duality of light?

Young's Experiment is one of the key experiments that demonstrated the wave-particle duality of light. It showed that light can behave as both a wave and a particle, depending on the conditions of the experiment. This led to the development of the quantum theory of light.

4. Can Young's Experiment be used to study other waves besides light?

Yes, Young's Experiment can be used to study other types of waves, such as sound waves and water waves. In fact, Thomas Young originally performed the experiment using water waves to demonstrate the wave-like behavior of light.

5. What are some real-world applications of Young's Experiment?

Young's Experiment has many practical applications, including the development of technologies such as diffraction gratings, which are used in spectroscopy to analyze the composition of materials. It has also been used in the field of quantum mechanics to study the behavior of particles at the quantum level.

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