Youngs and single slit experiments

AI Thread Summary
In Young's double slit experiment, decreasing the distance between the slits increases the distance between interference fringes, as confirmed by the formula sin(θ) = m(λ) / d. In a single slit diffraction experiment, increasing the width of the slit results in a wider central maximum on the screen due to enhanced light diffraction. Both conclusions align with established principles of wave optics. The discussion emphasizes the importance of understanding the mathematical relationships in these experiments. Overall, the explanations clarify key concepts in interference and diffraction phenomena.
Dx
Hiya!

I have questions that i am hoping someone can confirm my answers to, please.

1) In youngs double split experiment, if the separation between the slits decreased wouldn't the distance increase between the interference fringes? I say yes, i used this formula to derive my conclusion. sin[the]= m[lamb] / d. am i correct?

2) In a single slit diffraction experiment, if the width of the slit increases wouldn't the width of the central maximum on a screen increase as well? Is this true, from the pics and what I've read i believe it is but wanted a knowledgeable opinoin on this question
 
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Dx,

Homework problems go in the Homework Help forum, OK? These questions ain't theoretical physics.

Originally posted by Dx
1) In youngs double split experiment, if the separation between the slits decreased wouldn't the distance increase between the interference fringes? I say yes, i used this formula to derive my conclusion. sin[the]= m[lamb] / d. am i correct?

Yes.

2) In a single slit diffraction experiment, if the width of the slit increases wouldn't the width of the central maximum on a screen increase as well? Is this true, from the pics and what I've read i believe it is but wanted a knowledgeable opinoin on this question

Look at the formula for the minima and try again.
 
.

Hi there!

1) Yes, you are correct. In Young's double slit experiment, if the distance between the slits decreases, the distance between the interference fringes will increase. This can be seen from the formula you used, which relates the distance between the slits (d) to the wavelength of light (λ) and the angle of diffraction (θ). As the separation between the slits decreases, the angle of diffraction increases, resulting in a larger distance between the interference fringes.

2) You are also correct in your understanding of the single slit diffraction experiment. As the width of the slit increases, the width of the central maximum on a screen will also increase. This is because the wider slit allows for more diffraction of light, resulting in a broader central maximum. This can be observed from the diffraction pattern on a screen, where the central maximum will appear wider with a wider slit.

I hope this helps clarify your understanding of these experiments. Keep up the good work!
 
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