What is the angular width of the central maximum?

AI Thread Summary
The discussion centers on calculating the angular width of the central maximum for a single slit diffraction problem. A single slit of width 1.50 µm is illuminated with light of wavelength 500 nm, leading to confusion over the correct calculation method. The correct approach involves using the formula for the angular width, which requires the sine function and the inverse sine to find the angle. Participants clarify that the initial calculation mistake stemmed from misinterpreting the wavelength value, and emphasize the need to convert units correctly. Ultimately, the correct answer is confirmed to be 38.9 degrees after resolving the calculation errors.
wayneo
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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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Please show some work or initial thoughts.

~H
 
the answer in the book is 38.9 degrees but how is this right?
 
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?
 
anymore ideas anyone ?
 
The answer in the book is correct. You are simply a factor of 10^{2} out with respect to the \sin\theta. Your error lies here:

wayneo said:
(5 * 10^-9) / (1.5 *10^-6)

500nm \neq 5\times 10^{-9}m

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

~H
 
Last edited:
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
 
thanks guys it works, just made a silly error!
 
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