# Video: Determining the Wavelength of Light in a Double-Slit Interference Pattern

Find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m away from two slits separated by 0.120 mm.

02:30

### Video Transcript

Find the wavelength of light that produces fringes 7.50 millimeters apart on a screen 2.00 meters away from two slits separated by 0.120 millimeters.

If we draw in the diffraction pattern thatโs created by these Youngโs double-slit setup, we can see that in a vertical direction the spacing between the fringes โ that is the maxima โ is a constant value. If we take the vertical distance from the central line to the very first fringe above the central maximum and call that distance ๐ฆ, then that height ๐ฆ can form the opposite side of a right triangle.

Between the adjacent side and the hypotenuse, weโve defined an angle called ๐. Defining this angle ๐ is helpful because for a double-slit setup, there is a mathematical relationship that says that for the very first maximum fringe โ the one weโre focusing on โ ๐ the separation distance between the two slits times the sine of that angle ๐ is equal to the wavelength of the light used to create that fringe.

This relationship does only hold for the first fringe. But since thatโs the one weโre considering, we can work with it. So we can say that ๐ the separation distance between the slits which is given to us times the sine of the angle ๐ is equal to the wavelength ๐.

Looking at the geometry of our setup, we can see that this angle ๐ is a very small one. That is ๐ฟ is very large compared to ๐ฆ. When the angle ๐ is very small, that implies that the side of that angle is approximately equal to the tangent of that angle. That is the adjacent side, in this case ๐ฟ, is approximately equal in length to the hypotenuse. And we see that for our case, thatโs true.

As we consider the tangent now of the angle ๐, we can see geometrically that thatโs equal to ๐ฆ divided by ๐ฟ. We can now take this fraction and insert it in place of sin ๐ in our original expression.

Looking at this expression, weโre told in the problem statement what ๐ is: 0.120 millimeters, what ๐ฆ is: 7.50 millimeters, and what ๐ฟ is: 2.00 meters. To calculate the wavelength ๐ then, we only need to insert these values converted into units of meters and calculate this fraction.

When we do, we find a result of 4.50 times 10 to the negative seven meters or 450 nanometers. Thatโs the wavelength of the light that produces these fringes.