Question

Light with a wavelength of 550 nm is incident on a single slit, creating a diffraction pattern on a distant screen 2.5 m away. At a point on the diffraction pattern that is 1.443 m from the central maximum, the path length difference between the ray from the top of the slit and the bottom of the slit results in a phase difference of 6π radians. What is the width of the slit? (a) 1.1 µm (b) 2.2 µm (c) 3.3 µm (d) 4.5 µm (e) 5.3 µm

Answer 1

Solution:-

Given –

Wavelength (λ) – 550nm = 550*10^-9 m

Distance (L) – 2.5 m

Central maximum (D) – 1.443 m

The single slit diffraction pattern is given by,

I(ϴ) = I0sin²*(dsinϴ)/λ

Where the angle is given by,

sinϴ = x/L

Where L is the distance from the slit to the observation screen and x is the position in the observation screen referenced at the maximum intensity position, that is (x=0) = I0.thus the intensity distribution is given by,

I(x) = I0sin² (dx/L/λ) = 1

The first zero of the sine function intensity distribution occurs at,

dx0/L/λ = 1

THUS X0 is

d = λL/x0 = λL/D/2

Where,

D is the width of the central maximum then,

d = 0.00000055*2.5/1.443/2

d = 1.375*10^-6 / 0.7215

d = 1.90*10^-6

d = 1.1um

a)      1.1 um option is correct answer.