Light waves with two different wavelengths, 632 nm and 474 nm, pass simultaneously through a single slit whose width is 4.47 x 10-5 m and strike a screen 1.70 m from the slit. Two diffraction patterns are formed on the screen. What is the distance (in cm) between the common center of the diffraction patterns and the first occurrence of the spot where a dark fringe from one pattern falls on top of a dark fringe from the other pattern?

Answer 1

Hint: The wavelengths are in a nearly exact ratio of 4/3. The red light diffracts away from the center at a 33% faster rate.

The diffraction angles for maxima and minima can be found at
http://www.walter-fendt.de/ph14e/singleslit.htm

Consider the k=2 minimum for the ë1 = 474 nm light. It will occur where
sin á = 2 * ë1 /b
where b is the slit width.

The k=1 maximum for ë2 = 632 nm light will appear at
sin á = (3/2)* ë2/b

Since ë2/ë1 = 3/4,
these will be the same positions.
There will be other locations where the maximum of one color pattern coincides with the minimum of another, but this will be the closest to the central fringe.m

Answer 2

I was unable to type lambda (which appears as ë) and alpha (which appears as á ). I hope you can make it out.