Interference in a Double-Slit Experiment

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SUMMARY

The double-slit experiment discussed involves slits spaced 5.0mm apart and positioned 1.0m from the screen, producing interference patterns for light wavelengths of 480nm and 600nm. To find the separation between the third-order (m=3) bright fringes of these two wavelengths, the equation $d\sin\theta=m\lambda$ is utilized. The position of the bright fringes on the screen can be calculated using the relationship $OP=D\sin\theta$. This analysis allows for the determination of the fringe separation based on the specified parameters.

PREREQUISITES
  • Understanding of wave interference principles
  • Familiarity with the double-slit experiment setup
  • Knowledge of trigonometric functions in physics
  • Ability to manipulate equations involving wavelength and slit separation
NEXT STEPS
  • Calculate the angle θ for m=3 using the equation $d\sin\theta=m\lambda$ for both wavelengths
  • Determine the positions of the bright fringes on the screen using $OP=D\sin\theta$
  • Analyze the difference in fringe separation for varying wavelengths
  • Explore the implications of interference patterns in advanced optics
USEFUL FOR

Physics students, educators, and researchers interested in wave mechanics and optical experiments will benefit from this discussion.

annikaw
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In a double-slit experiment, the distance between slits is 5.0mm and the slits are 1.0m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 480nm, and the other due to light of wavelength 600nm. What is the separation on the screen between the third-order (m=3) bright fringes of the two interference patterns?

From my notes, I find that $d\sin\theta=m\lambda$.

However, I do not know how to get values of $\theta$ that can be put to the equation to find the difference in separation.
 
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annikaw said:
In a double-slit experiment, the distance between slits is 5.0mm and the slits are 1.0m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 480nm, and the other due to light of wavelength 600nm. What is the separation on the screen between the third-order (m=3) bright fringes of the two interference patterns?

From my notes, I find that $d\sin\theta=m\lambda$.

However, I do not know how to get values of $\theta$ that can be put to the equation to find the difference in separation.

Hi annikaw! Welcome to MHB! (Smile)

Let's get a drawing of the situation:
View attachment 4506

We get a bright fringe when $d\sin\theta=m\lambda$.
The corresponding position on the screen is $OP=D\sin\theta$.

Can you deduce where the 2 requested fringes are?
 

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