# Video: Slit-to-Screen Separation Required for a Given Intensity Minima Separation

A single slit of width 0.20 mm is illuminated by light of 400 nm wavelength. The diffracted light falls on a screen. In the pattern formed on the screen, the second minimum of the diffracted light is a distance of 2.5 mm from the central maximum. What is the distance between the slit and the screen?

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### Video Transcript

A single slit of width 0.20 millimeters is illuminated by light of 400-nanometer wavelength. The diffracted light falls on a screen. In the pattern formed on the screen, the second minimum of the diffracted light is a distance of 2.5 millimeters from the central maximum. What is the distance between the slit and the screen?

Let’s highlight some of the important information we’ve been given. We’re told that the width of the slit is 0.20 millimeters; we’ll call that value 𝑤. We’re also told that the wavelength of light on the slit is 400 nanometers; we’ll call that value 𝜆. And we’re told that the distance on the screen the light is projected on between the second minimum and the central maximum is 2.5 millimeters, a distance we’ll call ℎ.

We want to solve for the distance between the slit and the screen, which we’ll call capital 𝐷. Let’s start our solution by drawing a sketch of what’s going on. In this experiment, we have a single slit with light of wavelength 400 nanometers passing through it. The plane waves diffract through the slit and interfere with one another to create the interference pattern we see on the right side of the diagram on a screen.

We’re told that that vertical distance between the central maximum and the second minimum on the screen is ℎ. We want to solve for the distance between the slit and the screen 𝐷. For single-slit interference patterns, the distance on the screen ℎ between the central maximum and a minimum is approximated by an order number 𝑚 times the wavelength 𝜆 times the distance between the slit to screen 𝐷 all divided by the slit width 𝑤.

In our case 𝑚 equals two, so ℎ is equal to two 𝜆𝐷 over 𝑤. Rearranging this for 𝐷, we find that it’s equal to ℎ times 𝑤 over two 𝜆. Since we know each of these three values, we can plug them in now, being careful to write them in units of meters. When we enter these values on our calculator, we find that 𝐷 equals 0.63 meters. That’s the distance between the slit and the screen on which the interference pattern forms.