Question

A beam of monochromatic green light is diffracted by a slit of width 0.500 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 5.10 mm. Calculate the wavelength of the light.

Answer 1

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 issin(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 5.10mm. 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)=2.55/2060=12.378e-4. The wavelength=sin(q)*a=12.378e-4*.5=6.189e-4mm or 618.9nm.