Single Slit Diffraction - Find
Wavelength

## Question:

A beam of green light is diffracted by a slit of width 0.550 mm.
The diffraction pattern forms on a wall 2.06 m beyond the slit.
The distance between the positions of zero intensity on both
sides of the central bright fringe is 4.10 mm. Calculate the
wavelength of the laser light in nm.
## Answer:

The condition for zero intensity is that the waves from different
parts of the slit interfere destructively. This occurs when the
pathlength differs by 1/2 wavelength. The pathlength difference
between waves from the left half of the slit and those from the
right half is a/2*sin(q) where a is
the slit width and q is the
observation angle. If we set a*sin(q)/2=wavelength/2 then we find that
the condition for the first dark band is
sin(q)=wavelength/a. The second
dark band occurs at sin(q)=2*wavelength/a and so on.
The distance between adjacent dark bands on either side of the
central bright image is given as 4.10mm. The distance from the
center of the image to the first dark band would be 1/2 this
value. The wavelength then in terms of theta and a is wavelength=a*sin(q).
Putting in the values from the problem, making use of the fact
that the sine of small angles is about equal to the tangent, we
find sin(q)=2.05/2060=9.951e-4. The wavelength=sin(q)*a=9.952e-4*.55=5.473e-4mm or 547.3nm.