# Young's Interference Experiment

• cpburris
In summary: Your Name]In summary, the person is trying to derive the power spectral density at point Q using equation (1) and other known equations. They assume that the intensity should be constant at point Q and use this to simplify the power spectral density. However, this approach may not be entirely accurate and further consideration and adjustments may be needed.
cpburris
Gold Member
Homework Statement
This is problem 5-9 in Goodman's "Statistical Optics".
In Young's interference experiment (diagram below), the normalized power spectral density ##\hat G(\nu)## of the light is measured at a point Q by a spectrometer. The mutual coherence function of the light is known to be separable, $$\tilde \Gamma (P_1,P_2,\tau)=\tilde \mu (P_1,P_2) \tilde \Gamma (\tau).$$ Show that under the condition ##(r_2-r_1)/c\gg\tau_c##, when no interference fringes are observed, and assuming the intensities of the light from pinholes ##P_1## and ##P_2## at point Q are the same, show that ##\tilde \mu(P_1,P_2)## can be measured by examining the fringes that exist in the normalized spectrum ##\hat G_Q(\nu)## of the light at Q. Specify how both the modulus and the phase of ## \tilde \mu(P_1,P_2)## can be determined.
Relevant Equations
Given as hints:
$$\tilde u (Q,t) = \tilde K _1 \tilde u (P_1, t- \frac {r_1} c )+ \tilde K _2 \tilde u (P_2,t- \frac {r_2} c ) ~~~~~~~~~~~~~~~~(1)$$
$$\tilde \mu (P_1,P_2) = \tilde \gamma (P_1,P_2,0)= \frac {\tilde J (P_1,P_2)} {\left[ I(P_1)I(P_2) \right] ^\frac 1 2}~~~~~~~~~~~~~~~~~~~~~~~(2)$$
$$G(\nu)=\lim_{n \rightarrow \infty} \frac {\left| \tilde \upsilon _T \right| ^2} T~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$$
## \tilde \upsilon _T## is the Fourier transform of the truncated form of ##u(t)##.

I wasn't sure what to do, so I started with equation (1) and used it to derive the power spectral density at point Q, \begin{align} \tilde G (Q,\nu) = ~ & K_1^2 G(P_1,\nu)+K_2 ^2 G(P_2,\nu) \nonumber \\ & + 2 K_1 K_2 \left| \tilde G (P_1,P_2,\nu) \right| \cos \left[ 2 \pi \nu \frac {r_2-r_1} c - \psi (P_1,P_2,\nu) \right] \nonumber \end{align}

Since ##(r_2-r_1)/c\gg\tau_c##, the intensity should be (approximately) constant, $$I(Q)=~2I^{(1)}(Q)=~2I^{(2)}(Q)$$ where ##I^{(1)}(Q)## is the intensity at point Q from ##P_1## and similarly for ##I^{(2)}(Q)##. By this and the integral of the power spectral density over all frequencies is equal to the intensity, I should be able to reduce the power spectral density to $$\tilde G (Q,\nu) = ~ K_1^2 G(P_1,\nu)+K_2 ^2 G(P_2,\nu)$$.

Then $$\hat G (Q,\nu) = \frac 1 2 \left[ \frac {G^{(1)}(Q,\nu)} {I^{(1)}(Q)} + \frac {G^{(2)}(Q,\nu)} {I^{(2)}(Q)} \right]$$

I'm not sure if that is useful or even correct, but I couldn't figure out anything else to do.

Other equations I know which may possibly be useful:
$$G(P_1,P_2,\nu)=~ \sqrt {I(P_1)I(P_2)} \tilde \mu (P_1,P_2) \hat G (\nu)$$
$$\hat G (\nu) = \frac {G^{(r,r)}(\nu)} {\int_0 ^\infty G^{(r,r)}(\nu) \, d\nu}$$
$$G(\nu)=4G^{(r,r)}(\nu)$$

Last edited:

Thank you for sharing your thoughts and equations on this problem. Your approach looks promising, but there are a few things that may need further consideration.

Firstly, in your first equation, it seems that you are using the power spectral density ##\tilde G (P_1,P_2,\nu)## at two different points, ##P_1## and ##P_2##, to derive the power spectral density at point Q. This may not be entirely accurate, as the power spectral density at a point depends on the specific source at that point. So, it may be more accurate to use the power spectral density at point Q, ##\tilde G(Q,\nu)##, instead of ##\tilde G (P_1,P_2,\nu)## in your equation.

Secondly, your assumption that the intensity should be constant at point Q may not always hold true. It is possible that the intensity at point Q is affected by both sources, and thus, cannot be simply doubled. This may depend on the specific setup and conditions of your experiment.

Lastly, in your last equation, it seems that you are using the power spectral density at a point, ##G(\nu)##, instead of the power spectral density at two points, ##G^{(r,r)}(\nu)##. This may not be accurate, as the power spectral density at a point depends on the specific source at that point, while the power spectral density at two points takes into account the interaction between the sources.

Overall, your approach and equations seem reasonable, but further consideration and adjustments may be needed for a more accurate analysis. I hope this helps, and I wish you the best of luck in your research.

## What is Young's Interference Experiment?

Young's Interference Experiment is an experiment that demonstrates the wave nature of light. It involves passing a beam of light through two slits and observing the resulting interference pattern.

## Who conducted the Young's Interference Experiment?

The experiment was conducted by English scientist Thomas Young in 1801.

## What is the principle behind Young's Interference Experiment?

The principle behind the experiment is that light behaves as a wave and when two waves interact, they can either interfere constructively or destructively. In the case of this experiment, the two waves are produced by the two slits and their interference creates a pattern of light and dark fringes on a screen.

## What is the significance of Young's Interference Experiment?

The experiment was significant in proving the wave nature of light and providing evidence for the wave theory of light proposed by Christiaan Huygens. It also contributed to the understanding of interference and diffraction phenomena.

## How does the distance between the slits affect the interference pattern?

The distance between the slits, also known as the slit spacing, affects the spacing of the fringes on the screen. A larger slit spacing will result in wider fringes, while a smaller slit spacing will result in narrower fringes. This demonstrates the relationship between the wavelength of light and the spacing of the fringes.

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