# Waves & Optics: Bright Fringe Width Calculation

• eutopia
In summary: 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.The longest wavelength of visible light that would produce a dark fringe at the same location is 750 nm.
eutopia
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?

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.

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!

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

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?

yup hehehe... silly me

eutopia said:
yup hehehe... silly me
Good good.

## What is the equation for calculating the bright fringe width in a wave interference pattern?

The equation is given by w = λD/d, where w is the bright fringe width, λ is the wavelength of the light, D is the distance between the two slits, and d is the distance from the slits to the screen.

## How does the distance between the slits affect the bright fringe width?

The distance between the slits, D, is directly proportional to the bright fringe width, w. This means that as the distance between the slits increases, the bright fringe width will also increase.

## What happens to the bright fringe width when the distance between the slits is larger than the wavelength of the light?

If the distance between the slits is larger than the wavelength of the light, the bright fringe width will also be larger than the wavelength. This will result in a wider interference pattern with more distinct bright fringes.

## Can the bright fringe width be measured experimentally?

Yes, the bright fringe width can be measured experimentally by using a ruler or caliper to measure the distance between adjacent bright fringes on a screen. This measurement can then be compared to the calculated value using the equation w = λD/d.

## What factors can affect the accuracy of the bright fringe width calculation?

The accuracy of the bright fringe width calculation can be affected by factors such as the accuracy of the measurements of λ, D, and d, as well as external factors like the temperature and humidity of the environment. Additionally, any errors in the experimental setup or equipment can also affect the accuracy of the calculation.

• Classical Physics
Replies
4
Views
653
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
6K
• Introductory Physics Homework Help
Replies
14
Views
5K
• Introductory Physics Homework Help
Replies
2
Views
10K
• Introductory Physics Homework Help
Replies
2
Views
9K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
2K