Maximum Slit width diffraction minima

AI Thread Summary
The discussion focuses on determining the maximum slit width for which no visible light exhibits a diffraction minimum, with visible light wavelengths ranging from 400 nm to 750 nm. The key equation used is D*sin(theta) = m*lambda, where D is the slit width and lambda is the wavelength. Participants agree that when the slit width (D) is less than the smallest wavelength (400 nm), diffraction minima do not occur. The consensus concludes that a slit width of 400 nm is the maximum for avoiding diffraction minima across the visible spectrum. This understanding clarifies the relationship between slit width and wavelength in diffraction phenomena.
Decelerate
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Homework Statement



What is the maximum slit width so that no visible light exhibits a diffraction minimum?(Visible light has wavelengths from 400 nm to 750 nm.)


Homework Equations



D*sin(theta)=m*lambda ; where D = slit width,theta in degrees, m =1,2,3,..., and lambda is wavelength. The maximum slit width occurs at sin(90) =1 and if we take m=1 the equation reduces to D=lambda. When the slit width is equal to the wavelength of light passing through it then no diffraction minima are observed.

The Attempt at a Solution



For visible light, I just assumed that the maximum slit width had to be equal to the largest wavelength contained in visible light (750nm), but this does not make sense because then the wavelength at 450nm would still be defracted. I know for a fact that a slit width of 750nm is wrong, so I am guessing that the answer might be 450nm or maybe the average wavelength of visible light but am not sure

Any help is greatly appreciated.
 
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Minimum intensity occurs at \sin\theta_{min} = \frac{\lambda}{d} .

So when is there no \theta_{min}?
 
I guess my question is what wavelength do you use for visible light? It is a mixture of different wavelengths so How do I determine D when there are multiple wavelengths of light is it just the smallest wavelength (400) or the average...?
 
Decelerate said:
I guess my question is what wavelength do you use for visible light? It is a mixture of different wavelengths so How do I determine D when there are multiple wavelengths of light is it just the smallest wavelength (400) or the average...?

When does \theta_{min} not exist?
 
zachzach said:
When does \theta_{min} not exist?

When lambda is larger than d...I still don't really understand how that will help me determine the width for visible light from 400-750
 
Decelerate said:
When lambda is larger than d...I still don't really understand how that will help me determine the width for visible light from 400-750

So when d < \lambda , \theta_{min} DNE. So if d = 700nm what range does not have a minimum? What about if d = 600nm?
 
I had to edit that fyi.
 
so i guess d has to be 400nm for the range of 400-750 to not have a minimum
 
Decelerate said:
so i guess d has to be 400nm for the range of 400-750 to not have a minimum

That is my guess as well, less than 400nm but yeah.
 
  • #10
Let me enter it in and see if it is right.
EDIT: yes 400nm is right
Thanks!
 
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