Waves & Optics: Bright Fringe Width Calculation

AI Thread Summary
The discussion revolves around calculating the width of the bright fringe in a diffraction pattern created by light passing through a slit. A formula involving the wavelength, slit width, and distance to the screen is highlighted for this calculation. Additionally, participants discuss the speed of sound and electromagnetic radiation in relation to hearing an electric guitar from different distances. Clarifications on the equations and assumptions about the speed of light are made, with one participant expressing gratitude for the assistance. The conversation concludes with a successful resolution of the initial queries.
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 \frac{\lambda}{d} = \frac{x}{L} when \lambda 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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