When violet light of wavelength 415 nm falls on a single slit, it creates a central diffraction peak that is 8.80 cm wide on a screen that is 2.75 m away. How wide is the slit?

Answer 1

Is there something wrong with this? http://www.jiskha.com/display.cgi?id=1335108759 The part poorly defined is the "peak that is ...wide". The formula I gave you is the distance to the first minimum.
Some folks use the "width " of the peak to not the first minimum, but to the point where the peak has went down to the .707 RMS value.

Answer 2

I skipped over this question because I was not sure what was meant by width of the stripe. I think first minimum is most common.

Answer 3

The width of the central peak (max) is the distance between the first minima.
The equation for diffraction minima is
b•sinφ = k•λ.
k =1,
b = λ/sinφ.
From the geometry of diffraction pattern
tanφ =x1/L,
where x1 =8.8/2 = 4.4 cm =0.044 m is the distance between the center of diffraction pattern and the first min,
and L =2.75 m.
tan φ = 0.044/2.75=0.016.
sin φ = tan φ=0.016.
b = λ/sinφ = 415•10^-9/0.016 =
=2.59•10^-5 m =25.9 micrometers.

Answer 4

Thank you so much Elena! I forgot to divide 8.8 by two and, of course, kept getting double the answer which I knew was incorrect.

Answer 5

Anyone else from SF have to google this question, or was i the only one?

Answer 6

After 10 years and the question :
When violet light of wavelength 415 nm falls on a single slit, it creates a central diffraction
fringe that is 9.2 cm wide on a screen that is 2.55 m away. How wide is the slit?
Lmao still the same wording tho i dont understand it now

Answer 7

oh now i get it thanks elena