How far behind the pinhole should you place the viewing screen?

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Homework Help Overview

The problem involves determining the appropriate distance to place a viewing screen behind a pinhole to capture a circular diffraction pattern from a helium-neon laser. The context includes parameters such as the wavelength of the laser and the diameter of the pinhole.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the angular width of the central maximum and the parameters of the setup, including the wavelength and pinhole diameter. There are questions about which measurements to use and how to convert units appropriately.

Discussion Status

Participants are actively exploring different approaches to the problem, including setting up equations based on the diffraction pattern. Some have attempted calculations but are encountering difficulties with their results. There is no explicit consensus on the correct method or outcome yet.

Contextual Notes

There is a noted uncertainty regarding the appropriate values to use in calculations, particularly whether to use the diameter of the central maximum or the pinhole diameter. Participants are also converting units to ensure consistency.

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Homework Statement


You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0 cm. You have a helium-neon laser (lambda=633 nm) and a 0.12-mm-diameter pinhole.
How far behind the pinhole should you place the viewing screen?

Homework Equations





The Attempt at a Solution



I'm not understanding how to go about doing this/starting this question
 
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Half angular width of the central maximum = lambda/d, where d is the slit width.
And half angle = radius of the central maximum/distance of the screen.
 
so I do 633 nm/1 cm --> of course I will convert to uniform units
and half angle is 0.5 cm/x

but how do I use these?
and should i be using the 1 cm or the .12 mm?
 
633*x^-9/0.12*10^-3 = 0.5*10^-2/x
 
Last edited:
so I did that and it's not right. so then I tried using 0.12 x 10^-3 instead of 0.06x 10^3
and I solved for x, and it's not right...
 
x = 0.5*10^-2*0.12*10^-3/633*10^-9 = 0.95m
 

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