Single-slit diffraction diffraction pattern

  • Thread starter drawar
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  • #1
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Homework Statement


A single slit forms a diffraction pattern with monochromatic light. The 6th minimum of the
pattern occurs at an angle of 23° from the central maximum. The number of bright bands on
either side of the central band is closest to:
A) 16 B) 13 C) 14 D) 15 E) 17


Homework Equations


asin(theta)=mlambda


The Attempt at a Solution


Honestly I have no idea how to do this. Using the data given, I can only set up one equation with 2 unknowns, which is impossible to solve. On top of that I am unable to deduce a relationship between maxima and minima in a single-slit diffraction. I hope someone would throw me some light on this, thanks!
 

Answers and Replies

  • #2
PeterO
Homework Helper
2,434
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Homework Statement


A single slit forms a diffraction pattern with monochromatic light. The 6th minimum of the
pattern occurs at an angle of 23° from the central maximum. The number of bright bands on
either side of the central band is closest to:
A) 16 B) 13 C) 14 D) 15 E) 17


Homework Equations


asin(theta)=mlambda


The Attempt at a Solution


Honestly I have no idea how to do this. Using the data given, I can only set up one equation with 2 unknowns, which is impossible to solve. On top of that I am unable to deduce a relationship between maxima and minima in a single-slit diffraction. I hope someone would throw me some light on this, thanks!

There is a formula governing the angle to each minimum which is something like

sin θ = (n) λ/W

where is the wavelength of light, and W is the width of the slit, and n is a count of the number of the minimum - so would be 6 here; for the 6th minimum.

Your formula looks like a transformation of that to W sin θ = (n)λ

Anyhow, once you have established the value of λ/W, you can work out the maximum value of (n) - since the sinθ has a maximum value of 1.
 

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