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What is the width? Please take a look at my work!

  • Thread starter ixerr
  • Start date
  • #1
24
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Homework Statement


So I have tried to solve this problem but my answer keeps turning out to be wrong.. Someone please tell me what I am doing wrong?
A 0.484 mm diameter hole is illuminated by light of wavelength 556.0 nm. What is the width (in mm) of the central maximum on a screen 3.45 m behind the slit?

Homework Equations


The width of the central maximum is given by W= (2λL)/(a)
So I have wavelength λ=556x10^-9 m
Screen distance L=3.45 m
And slit width= .484x10^-3 m

The Attempt at a Solution


So the width of the central maximum = ((2)(556x10^-9)(3.45))/(0.484x10^-3) = 7.926 mm
 

Answers and Replies

  • #2
349
1
I get the same answer as you and I can't see anything that you have done wrong.
What is the answer you have been given? is it half your calculated answer or does it include the diameter of the original aperture (7.9 + 0.484mm)
 
  • #3
24
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There is no answer.. It's an online homework thing, and it keeps saying I have the incorrect answer. I am at a loss here:(
 
  • #4
349
1
Well that is annoying !!! I got angular width Sinθ = λ/a = 556x10^-9/0.484x10^-3 = 1.149x10^-3
I then did calculation assuming small angles (which is excellent for this angle) BUT I ignored the 0.484mm slit width.....to get width from w/3.45 = 1.149x10^-3 which gives 3.96 x 10^-3m (3.96mm) this means the total width of the central max = 2 x 3.96 = 7.92mm
I think this could be 7.92 + 0.484 = 8.41mm if you had to include the width of the slit.
Wish I knew what answer you were expected to produce...!!! hope this is some help.
You have certainly used the correct method.
 
  • #5
349
1
I have just realised one more thing ....Sin∅ = 1.22λ/a for a circular aperture
Bad mental block on my part to miss that....see what difference that makes to your calculation
 
  • #6
24
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Yes, I got the answer now :) Thank you so much for all the help!
 

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