Question
Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of $45.0^\circ$
$290 \textrm{ nm}$
Solution Video

# OpenStax College Physics Solution, Chapter 27, Problem 9 (Problems & Exercises) (1:20) View sample solution

## Calculator Screenshots Video Transcript

This is College Physics Answers with Shaun Dychko. We begin by writing down the information we're given. The wavelength is 410 nanometers for the violet light that is going to be incident on these two slits. The angle to the first minimum is 45 degrees and when you're thinking about what value <i>M</i> should have in this formula here for the minima through a double slit, the first minimum is at <i>M</i> equals zero. It's not going to be one because it's not asking for the first order minimum. Instead, it's asking for the first minimum, which is <i>M</i> equals zero. So, the minima is going to be the distance between the slits times sine of the angle to the minimum. That equals the order of the minimum plus a half times the wavelength. So, we can solve for <i>D</i>, the distance between the slits, by dividing both sides by sine <i>Theta</i> and we get the distance between slits, then, is <i>M</i> plus a half times <i>Lambda</i> over sine <i>Theta</i>. So, that's zero plus a half times 410 nanometers divided by sine of 45. And because I left this in units of nanometers, our answer is also in units of nanometers and that will be 290 nanometers is the separation between the slits.

Submitted by juliasoncini on Wed, 04/15/2020 - 14:35

How did you know to use the destructive interference equation rather than the constructive interference equation?

Submitted by ShaunDychko on Sat, 04/18/2020 - 09:59

Hello juliasoncini, thanks for the question. This problem asks for the first minimum. This is the position on the screen where there will be almost no light at all. In other words, the brightness of the light at this position will be a minimum. This minimum occurs because waves from each of the two slits travel a different distance such that when they arrive at the screen they are perfectly out of phase. They interfere destructively. The amplitudes of their electromagnetic waves are opposite (one has positive maximum while the other has an equal magnitude minimum at this position on the screen). The waves destroy each other, so there is no brightness, and this is called a minimum. The destructive interference equation tells us how to find the positions of these minima.
All the best,
Shaun