Question

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.

Answer 1

a) 1.17 m

b) 0.929mm

Step-by-step explanation:

(a) to find the distance to the screen you use

`y=(m\lambda D)/(d)`

m: order of the fringe

lambda: wavelength of the light = 588*10^{-9} m

D: distance to the screen

d: distance between slits = 0.74*10^-3 m

by doing D the subject of the formula and replacing the values of the other parameters you obtain:

`D=(dy)/(m\lambda)=((0.74*10^(-3)m)(0.93*10^(-3)))/((1)(588*10^(-9)m))=1.17m`

the distance to the screen is 1.17m

(b) to find the width of the central maximum you calculate the position of the first dark fringe:

`y=((1)(588*10^(-9)m/2)(1.17m))/(0.74*10^(-3)m)=4.64*10^(-4)m=0.464mm`

`y=(m(\lambda/2)D)/(d)`

2y = 2(0.464mm)=0.929mm is the width of the central maximum

Answer 2

a) L = 1.17 m

b) width of central maxima = 1.28 mm

Step-by-step 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