Light of wavelength 588.0 nm illuminates a slit of width 0.74 mm. (a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.93 mm from the central maximum(b) Calculate the width of the central maximum.

a) L = 1.17 m

b) width of central maxima = 1.28 mm

Explanation:

Given:-

- The wavelength of light, λ = 588.0 nm

- The slit of width, a = 0.74 mm

Find:-

(a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.93 mm from the central maximum

(b) Calculate the width of the central maximum.

Solution:-

- The results of Young's single slit experiment are given in form of a relation as the angle of separation between fringes ( θ ) as function of fringe order ( m ) and wavelength ( λ ).

- Destructive interference produces the dark fringes ( minimum ).  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which:

                        a*sin ( θ ) = m*λ

Where,

            m : The order number for the minimum ( dark fringe ).

- We are to investigate for the first (m = 1 )) dark fringe (minima) which is y = 0.93 mm from central order ( m = 0 ) for which the screen must be placed at a distance L:

                        a*y/L = m*λ

                        L = a*y / m*λ      

                        L = (0.74*0.93) / (1*588*10^-9)    

                        L = 1.17 m

- The distance L of screen should be 1.17 m away from slit.

- The central maximum - central bright fringe. The maxima lie between the minima and the width of the central maximum is simply the distance between the 1st order minima from the centre of the screen on both sides of the centre.

                                tanθ ≈ θ ≈ y/L = w*λ

                                y = w*λ*L

The width of the central maximum is simply twice this value:

- Width of central maximum = 2λLw = 2*588*10^-6*1170*0.93  = 1.28 mm                                

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