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 that’s created by these Young’s double-slit
setup, we can see that in a vertical direction the spacing between the fringes —
that is the maxima — 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 𝑦, then that height 𝑦 can form the
opposite side of a right triangle.
Between the adjacent side and the hypotenuse, we’ve defined an angle called 𝜃. Defining this angle 𝜃 is helpful because for a double-slit setup, there is a
mathematical relationship that says that for the very first maximum fringe — the one
we’re focusing on — 𝑑 the separation distance between the two slits times the sine
of that angle 𝜃 is equal to the wavelength of the light used to create that
This relationship does only hold for the first fringe. But since that’s the one we’re considering, we can work with it. So we can say that 𝑑 the separation distance between the slits which is given to us
times the sine of the angle 𝜃 is equal to the wavelength 𝜆.
Looking at the geometry of our setup, we can see that this angle 𝜃 is a very small
one. That is 𝐿 is very large compared to 𝑦. When the angle 𝜃 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 𝐿, is approximately equal in length to the
hypotenuse. And we see that for our case, that’s true.
As we consider the tangent now of the angle 𝜃, we can see geometrically that that’s
equal to 𝑦 divided by 𝐿. We can now take this fraction and insert it in place of sin 𝜃 in our original
Looking at this expression, we’re told in the problem statement what 𝑑 is: 0.120
millimeters, what 𝑦 is: 7.50 millimeters, and what 𝐿 is: 2.00 meters. To calculate the wavelength 𝜆 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. That’s the wavelength of the light that produces these fringes.