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Single Slit Diffraction Intensity

  • Thread starter 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
 

Answers and Replies

  • #2
TSny
Homework Helper
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The Attempt at a Solution


(i)
we find a = 2.37375*10^-4 m
I believe that's correct

(ii)
Is it in correct to m+.5 for constructive interference?
No, using (m+.5)(lamda) = asin(theta) for the maxima is not very accurate. You should find the maxima by finding where the intensity function takes on its maxima. The first maximum beyond the central maximum does not occur at a small value of β/2, so you should not use the approximation for sin(β/2) in this part of the problem.

EDIT: Actually, using (m+.5)(lamda) = asin(theta) with m = 1 gives an answer for part (ii) to an accuracy of about 4.6%. So, it depends on how accurate you need to be here. The reason you were way off in your answer is because you used the approximation for sin(β/2), which is not accurate at all here.
 
Last edited:
  • #3
143
1
I believe that's correct


No, using (m+.5)(lamda) = asin(theta) for the maxima is not very accurate. You should find the maxima by finding where the intensity function takes on its maxima. The first maximum beyond the central maximum does not occur at a small value of β/2, so you should not use the approximation for sin(β/2) in this part of the problem.

EDIT: Actually, using (m+.5)(lamda) = asin(theta) with m = 1 gives an answer for part (ii) to an accuracy of about 4.6%. So, it depends on how accurate you need to be here. The reason you were way off in your answer is because you used the approximation for sin(β/2), which is not accurate at all here.
Thanks for the reply,
would it be more accurate to look for when the sin function is 1?
so for instance its first 1 when β = pi so
I = (I0)(1/(pi/2))^2 = (I0)*(4/pi^2)
 
  • #4
TSny
Homework Helper
Gold Member
12,403
2,839
Thanks for the reply,
would it be more accurate to look for when the sin function is 1?
so for instance its first 1 when β = pi so
I = (I0)(1/(pi/2))^2 = (I0)*(4/pi^2)
No. If you graph I = Io[sin(β/2)/(β/2)]2, you will see that β = ##\pi## is not near a maximum. In fact, the first minimum occurs at β = ##2\pi##. So, the first maximum beyond the first minimum occurs well beyond β = ##2\pi##.
 

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