Single Slit Diffraction Intensity

In summary: Therefore, using the approximation for sin(β/2) is not accurate here.The correct way to find the maxima is to set the derivative of the intensity function equal to 0 and solve for β. You will find that β = ##2.404825558## is the first maximum beyond the central maximum. This value of β is about 4.6% more than the value you calculated by using the approximation for sin(β/2). So, you could use that value of β along with the formula for I = Io[sin(β/2)/(β/2)]2 and get a result that is accurate to within about 4.6%.
  • #1
Woolyabyss
143
1

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
 
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  • #2
Woolyabyss said:

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
TSny said:
I believe that's correctNo, 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
Woolyabyss said:
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##.
 

Related to Single Slit Diffraction Intensity

What is single slit diffraction intensity?

Single slit diffraction intensity is the phenomenon where light passing through a narrow slit diffracts, or spreads out, creating a pattern of bright and dark fringes. This occurs due to the wave nature of light.

What factors affect single slit diffraction intensity?

The intensity of single slit diffraction is affected by the width of the slit, the wavelength of the light, and the distance between the slit and the screen. The narrower the slit, the wider the diffraction pattern will be. A shorter wavelength of light will also result in a wider diffraction pattern, while increasing the distance between the slit and the screen will decrease the intensity of the diffraction pattern.

How can single slit diffraction be used in real-life applications?

Single slit diffraction is commonly used in spectrometers, which are instruments used to measure the wavelengths of light. It is also used in the production of holograms and in optical tweezers, which use diffraction to manipulate small particles.

What is the mathematical equation for calculating single slit diffraction intensity?

The equation for calculating single slit diffraction intensity is I = I0 * (sin(πaλ/θ)/(πaλ/θ))^2, where I0 is the initial intensity of the light, a is the width of the slit, λ is the wavelength of the light, and θ is the angle of diffraction.

How does single slit diffraction differ from double slit diffraction?

Single slit diffraction produces a pattern with a central bright fringe and alternating dark and bright fringes on either side. Double slit diffraction, on the other hand, produces a pattern with multiple bright fringes separated by dark fringes. Single slit diffraction also has a narrower diffraction pattern compared to double slit diffraction, due to the single source of light passing through the single slit.

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