- #1

Woolyabyss

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## Homework Statement

A narrow slit is illuminated by a collimated 633-nm laser beam as shown in the figure below. This produces a diffraction pattern on a screen, which is 6 m away from the slit as shown in the figure below. The distance between the centres of the first minima outside the central bright fringe is 32 mm.

(i) Find the width of the slit.

(ii) Calculate the intensity of the secondary maximum relative to the intensity of the central maximum, I0, in the diffraction pattern.

(iii) Find the width of the central bright fringe measured at its half maximum intensity.

use the following approximation sin(β/2) = β/2 - (β/2)^3/6

## Homework Equations

m(lamda) = asin(theta) destructive interference

I = (I0)(sin(β/2)/(β/2))^2

## The Attempt at a Solution

(i)

m = 1

lamda = 633*10^-9 m

Y = 16*10^-3 m

sin(theta) = theta = tan(theta) ...(approximation for small angle)

Y/R = tan(theta)

rearranging and using sin(theta) = tan(theta)

Y = R(m)(lamda)/a

rearranging in terms of a

we find a = 2.37375*10^-4 m

(ii)

using the approximation

I = (I0)(sin(β/2)/(β/2))^2

becomes

I = (I0)(1-(β/2)^2/6)^2

for constructive interference

m(lamda) = asin(theta) becomes

(m+.5)(lamda) = asin(theta)

for m = 1

the phase difference must be 2*pi*(m+.5) = 3*pi for m = 1

subbing this into I = (I0)(1-(β/2)^2/6)^2

I=(I0)*7.295

I'm not sure where I wen't wrong. Is it in correct to m+.5 for constructive interference?

any help would be appreciated