Single-slit diffraction diffraction pattern

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A single slit diffraction pattern is analyzed with monochromatic light, where the 6th minimum occurs at an angle of 23° from the central maximum. The relevant equation is a sin(θ) = mλ, which relates the angle to the wavelength and slit width. A formula for the angle to each minimum is also discussed: sin(θ) = (n)λ/W, where n is the minimum count. By determining the ratio of wavelength to slit width (λ/W), one can calculate the maximum number of bright bands on either side of the central band. The discussion emphasizes the need for clarity on the relationship between maxima and minima in single-slit diffraction patterns.
drawar
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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!
 
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drawar said:

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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