Question

Width of Peak (cm): 1.6 Width of Slit for Single Peak: 2.3

The width of the center peak (w) can be calculated using the equation w = 2λL/ a

where a is the width of the slit. Use this equation to determine the width of the slit that caused the diffraction pattern. Show your work.

I am confused as to what I am finding since I do not have L. I have going backwards λL = 1.84. If someone could explain what I should actually be doing in this question that would be amazing

Answer 1

The condition for dark fringe at either side of central bright fringe in single slit configuration is:

,

where a is slit width [it must be very less (of the order of mm) to see any appreciable width (w) of diffraction fringes at screen by visible light at L distance from the slit. n is the order of diffraction, i.e. number of dark fringe starting from the central bright fringe. is the wavelength of light, i.e. of the order of 100 nm.

To determine the width of central bright fringe n must be taken as 1. If L is very large (of the order of meters) then can be replaced by w/2L. The above equation will take the form as given in the question.

or

In the problem only a and 2w are given. So either L or may be asked.

If this experiment has been performed using visible light (any of wavelength from 380 - 740 nm) the distance of screen L comes out to be 484 - 248 m, which is not possible to bring in existence in laboratory environment.

Here I wish to bring your notice in to the fact that the given width of slit is very large (2.3 cm) and the width of peak (central bright fringe) is shorter that the slit itself. It is not the usual case. The purpose of this type of question may be to show you that the estimate of slit width has impact on applicability of the phenomenon in practical situations. Therefore I think you are going to find that how wrong a situation can be if the parameters are not chosen wisely.

I hope I've contributed to remove your confusion.