Waves & Optics: Bright Fringe Width Calculation

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Homework Help Overview

The discussion revolves around calculating the width of a bright fringe in a diffraction pattern created by light passing through a slit, as well as addressing a related sound propagation problem. The subject area includes waves and optics, particularly focusing on diffraction and interference patterns.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of specific equations related to diffraction, such as \(\frac{\lambda}{d} = \frac{x}{L}\), and question their familiarity with these concepts. There is also a consideration of the speed of electromagnetic radiation in relation to sound propagation.

Discussion Status

The discussion is active, with participants sharing equations and clarifying concepts. Some participants express uncertainty about the equations and seek confirmation on their assumptions regarding physical constants. There is a mix of attempts to solve the problems and clarifications being sought.

Contextual Notes

Participants are working within the constraints of homework rules, which may limit the information they can share or the methods they can use. There is also a mention of the range of visible wavelengths and the specific requirements for producing dark fringes in the context of Young's double slit experiment.

eutopia
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The width of a slit is 19.3 mm. Light with a wavelength of 505 nm passes through the slit and falls on a screen that is located 48.7 cm away. In the diffraction pattern, find the width of the bright fringe that is next to the central bright fringe.

Assuming that the speed of sound is 340 m/s, how far from the amplifier should you sit to hear the electric guitar at the same time as a person 2.95×107 meters away in a spaceship will hear the guitar listening with headphones to a live radio transmission?
 
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eutopia said:
The width of a slit is 19.3 mm. Light with a wavelength of 505 nm passes through the slit and falls on a screen that is located 48.7 cm away. In the diffraction pattern, find the width of the bright fringe that is next to the central bright fringe.
This one needs [tex]\frac{\lambda}{d} = \frac{x}{L}[/tex] when [tex]\lambda[/tex] is the wavelength, d is the diffration grating size, x is the fringe spacing and L is the distance the diffration grating is from the place of observation. Do you reckonise that equation at all or is it new to you?

eutopia said:
Assuming that the speed of sound is 340 m/s, how far from the amplifier should you sit to hear the electric guitar at the same time as a person 2.95×107 meters away in a spaceship will hear the guitar listening with headphones to a live radio transmission?
Am I to assume that 2.95 x 107 means 2.95 x 107?

All you need to do is work out how long it will take for the sound to reach the spaceship (assuming that the speed of electromagnetic radation is 3 x 106 ms-1). Then you can use that time, with the speed of sound, to work out your distance.

The Bob (2004 ©)
 
hey thanks! no, i didn't know that equation, the first one for the waves. i was trying to find an equation for that problem and couldn't find one anywhere in my book.

im assuming the speed of electromagnetic radiation is 3e8, right? and it worked out too.. thanks!
 
eutopia said:
hey thanks! no, i didn't know that equation, the first one for the waves. i was trying to find an equation for that problem and couldn't find one anywhere in my book.

im assuming the speed of electromagnetic radiation is 3e8, right? and it worked out too.. thanks!
Yes, electromagnetic radiation is 3e8.

Have you got the answers you wanted? If so good, no problem and I was happy to help. :smile:

The Bob (2004 ©)
 
In Young's double slit experiment, 425 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What is the longest wavelength of visible light that would produce a dark fringe at the same location? Assume that the range of visible wavelengths extends from 380 to 750 nm.

I did 425nm/4 times 3.5, but the computer says I'm wrong and I don't know why.
 
oh i got it! it says the LONGEST... so I had to divide by 2.5
 
So have you got the answers you wanted?

The Bob (2004 ©)
 
yup hehehe... silly me
 
eutopia said:
yup hehehe... silly me
Good good. :smile:

The Bob (2004 ©)
 

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