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What is the angular width of the central maximum?

  • Thread starter wayneo
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  • #1
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A single slit of width 1.50 um is illuminated with light of wavelenth 500.0 nm. What is the angualar width of the central maximum?
 

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  • #2
Hootenanny
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Please show some work or initial thoughts.

~H
 
  • #3
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the answer in the book is 38.9 degrees but how is this right?
 
  • #4
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I tried the formula &= wavelenth / b and came up with (5 * 10^-9) / (1.5 *10^-6) and got 3.33 * 10^-3 then multiplied this by 2. is this right?
 
  • #5
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anymore ideas anyone ???
 
  • #6
Hootenanny
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The answer in the book is correct. You are simply a factor of [itex]10^{2}[/itex] out with respect to the [itex]\sin\theta[/itex]. Your error lies here:

wayneo said:
(5 * 10^-9) / (1.5 *10^-6)
[tex]500nm \neq 5\times 10^{-9}m[/tex]

Recheck your calculation. Also you need to find the angle (ArcSin)

~H
 
Last edited:
  • #7
nrqed
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wayneo said:
I tried the formula &= wavelenth / b and came up with (5 * 10^-9) / (1.5 *10^-6) and got 3.33 * 10^-3 then multiplied this by 2. is this right?
first, 500 nm = 500 times 10^-9.

The second thing is that what you found is sin(theta). To find the *angle* you must do the inverse sin of that. Then multiply the result (in degrees) by two
 
  • #8
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thanks guys it works, just made a silly error!
 

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