A Fraunhofer diffraction pattern is produce on a screen

[AnnieMoo]

Homework Statement

A Fraunhofer diffraction pattern is produced on a screen 218 cm from a single slit. The distance from the center of the central maximum to the first-order maximum is 7620 times the wavelength (lambda). Calculate the slit width. Answer in __m

Homework Equations

The Attempt at a Solution

I have no idea how to even start this problem but this is what I tried.

7620 = (1.5 * 2.18)/a where a is the slit width.
This was not correct. I would really appreciate any kind of help on this!

 

[anathema]

Hello, thank you for posting your problem. I can help guide you through solving this problem.

The first thing we need to do is understand the Fraunhofer diffraction pattern and what it tells us. The Fraunhofer diffraction pattern is formed when a single slit is illuminated with a coherent light source, such as a laser. It produces a pattern of bright and dark fringes on a screen placed at a certain distance from the slit.

In this problem, we are given the distance from the screen to the slit (218 cm) and the distance from the central maximum to the first-order maximum (7620 times the wavelength). We also know that the distance between adjacent bright fringes is equal to the wavelength (lambda).

Now, let's think about what the distance between the central maximum and the first-order maximum represents. Since we know that the distance between adjacent bright fringes is equal to the wavelength, and the first-order maximum is 7620 times the wavelength away from the central maximum, this means that there are 7620 bright fringes between the central maximum and the first-order maximum.

So, we can set up an equation using this information:

7620 * lambda = distance between central maximum and first-order maximum
lambda = distance between adjacent bright fringes

Now, we can rearrange this equation to solve for the wavelength:

wavelength = (distance between central maximum and first-order maximum) / 7620

We know that the wavelength is equal to the slit width (a) times the distance between adjacent bright fringes, so we can set up another equation:

wavelength = a * (distance between adjacent bright fringes)

Now, we can combine these two equations and solve for the slit width (a):

a = (distance between central maximum and first-order maximum) / (7620 * distance between adjacent bright fringes)

Plugging in the values we were given, we get:

a = (218 cm) / (7620 * lambda)

Now, we need to convert the given distance in centimeters to meters, and we know that 1 m = 100 cm, so:

a = (2.18 m) / (7620 * lambda)

Finally, we can substitute the given value for the distance between the central maximum and first-order maximum (7620 * lambda) into this equation:

a = (2.18 m) / (7620 * 7620 * lambda)

We can simplify this to: