Find the angular width of the central maximum.

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Homework Help Overview

The discussion revolves around calculating the angular width of the central maximum in a single slit diffraction problem, specifically involving a slit width of 1.50 * 10-6 m and light of wavelength 500.0 nm.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the angular width, with one noting the formula θ = λ / b and questioning the interpretation of the result. Others explore the need to consider the total angular width (2Θ) and the implications of using radians versus degrees in their calculations.

Discussion Status

The discussion is active, with participants sharing their calculations and questioning discrepancies in results. Some guidance has been offered regarding the interpretation of angular positions and the use of different units, but no consensus has been reached on the exact values.

Contextual Notes

Participants mention potential issues with unit conversions and the validity of external resources, indicating a need for clarity on these points. There is also an acknowledgment of a fundamental misunderstanding that may be affecting calculations.

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Homework Statement


A single slit of width 1.50 * 10-6m is illuminated with light of wavelength 500.0 nm. Find the angular width of the central maximum.

Homework Equations


θ = λ / b

The Attempt at a Solution


b = 1.50 * 10-6m
λ = 5.000 * 10-7
θ = λ / b = 0.333 = 19.1°

But the answer is 38.9°.

All my calculation of diffraction on single slit is off-target. It seems there is a fundamental point that I am missing. Please help. :frown:
 
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You have found the angular position,measured from the centre,of the pattern.There are two such minima,one on each side of the central maxima.
 
Last edited:
OK.. so what I need is 2Θ, right? Then I got 38.2, and this is a bit off from 38.9. Where did this difference come from?
 
I calculated it to be 38.9.The equation is:
sin theta=lambda/d.
Did you take theta to be in radians?If so that is a good approximation but only for very small angles.
 
I had a similar problem and was given this link.
http://www.calctool.org/CALC/phys/optics/fNA
It is a great link and solved my problem with units
 
Now I can solve the problems. Thanks everyone! :D

p.s. Emily, the link is not valid. Could you give me a right one? I can solve the problems, but still I want to check it out.
 

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