# Intensity in a Double-Slit problem?

• Newtime
In summary, the conversation discusses the confusion over solving a problem related to an old exam. The equation I = I0cos^2(phi) is given and the goal is to double the intensity. It is determined that positive interference occurs at y = m*lamba*L/d, and the intensities add up to 2*I_0. It is later discovered that the correct answer is y = m*lamba*L/d divided by 4, and this is confirmed by plotting the intensity function. The initial confusion is resolved with the help of the correct equation.
Newtime
I've been looking over an old exam for one of my classes, and cannot decipher the train of thought used in solving (or method used) the problem on the last page:

http://people.physics.tamu.edu/kochar/FE/FE_Fall05_1.pdf

Specifically, for the first part, what is going on...? Firstly, I have the equation that I = I0cos^2(phi). So if I want the intensity doubled then I end up with cos^2(phi) = 2, which isn't possible? Clearly I am thinking about this in the entirely wrong way and could use some help, thanks in advance.

I think it means just where there is positive interference. Which would be approximately y = m*lamba*L/d. m=1,2,3...

The intensities add up, so if you get intensity I_0 from one beam and the same from another beam you get 2*I_0 at the point where there is maximum constructive interference.

Nevermind, I figured it out.

kcdodd said:
I think it means just where there is positive interference. Which would be approximately y = m*lamba*L/d. m=1,2,3...

That was my first thought, but turns out it was in correct. The answer happens to be what you posted above for m=1 but all divided by 4 since the intensity equals the sum of the individual intensities and the point one fourth of the way from the center point of the interference pattern to the center point of the first bright spot - the formula you gave. if you plot the intensity function, this becomes quite clear. but thank you for your reply.

## What is intensity in a Double-Slit problem?

Intensity in a Double-Slit problem refers to the amount of light or energy per unit area that is passing through the slits. It is a measure of the brightness or strength of the light waves.

## How is intensity related to the double-slit experiment?

In the double-slit experiment, light waves passing through two parallel slits interfere with each other, creating a pattern of bright and dark bands on a screen. The intensity of the light at each point on the screen is determined by the constructive or destructive interference of the waves.

## What factors affect the intensity in a Double-Slit problem?

The intensity in a Double-Slit problem is affected by the wavelength of the light, the distance between the slits, and the distance between the slits and the screen. It can also be affected by the properties of the material used for the slits and the screen.

## Why is intensity important in the double-slit experiment?

Intensity is important in the double-slit experiment because it helps us understand the nature of light as a wave and how it behaves when interacting with objects. It also allows us to make predictions and calculations about the interference patterns that will form on the screen.

## How is intensity measured in a Double-Slit problem?

Intensity is typically measured in units of power per unit area, such as watts per square meter. In the double-slit experiment, it can be calculated by measuring the brightness of the light at different points on the screen and using the equation for intensity: I = P/A, where I is intensity, P is power, and A is area.

Replies
2
Views
3K
Replies
3
Views
2K
Replies
17
Views
4K
Replies
9
Views
5K
Replies
5
Views
3K
Replies
7
Views
2K
Replies
4
Views
2K
Replies
105
Views
5K
Replies
8
Views
2K
Replies
5
Views
1K