The distance between the 3rd order bright spot and the centre of the central maximum is 12.0 cm. The screen is 2.0 m away. Determine the single slit width if wavelength is 550 nm.
When light is passed through a single slit and falls on a screen placed at a distance there is a formation of bands due to diffraction.
The following equations are used to determine the properties of the diffraction pattern. If the width of the slit is denoted by A, the wavelength of light passing through the slit is denoted by W, L stands for the distance of the screen from the slit and p denotes the number of the fringe, we have:
A*sin x = p*W, where p = 1,2,3...
If the distance of the pth maximum from the central maximum is y,
y= L*tan x
For small angles we can equate the values of the sine and the tangent functions.
This gives: y/L = p*W/A
In the problem we are given that p = 3, y = 12 cm. L = 2 m and W = 550 nm, we need to determine A.
A = p*W*L/y
=> A = 3*550*10^-9*2/12*10^-2
=> A = 275*10^-7 m
=> A = 0.00275 cm
The single slit width is equal to 0.00275 cm