Diffraction Grating interference pattern

AI Thread Summary
Replacing two slits with 20 slits of the same spacing will not change the number of bright fringes observed on the screen; the positions of these fringes remain the same. The interference pattern will still have the same number of maxima, but the bright fringes will be narrower and brighter due to increased slit count. The formula for constructive interference, dsin(theta) = m(lambda), applies equally to both configurations. Although the fringe spacing increases, the overall number of visible fringes does not change. Thus, the interference pattern's characteristics are enhanced without altering the number of fringes.
JJBladester
Gold Member
Messages
281
Reaction score
2

Homework Statement



This is the interference pattern on a viewing screen behind two slits.

(image of 7 bright fringes with black fringes between every pair of bright fringes)

How would the pattern change if the two slits were replaced by 20 slits having the same spacing d between adjacent slits? Would the number of fringes on the screen increase, decrease, or stay the same?

Homework Equations



dsin(theta)=m(lambda)

The Attempt at a Solution



The formula above says that for constructive interference, the path-length difference between successive slits needs to be an integer multiple of wavelength (lambda). However, I don't see anything in my book about the number of fringes going up or down with N (the number of slits) on the grating. Any advice?

I know that the fringe spacing increases, the width of each fringe decreases, and the brigntness of each fringe increases, but I don't know about the *number* of fringes and whether it goes up or down.
 
Physics news on Phys.org
ehild said:

Two-thirds of the way down, I see:

eq_diff_slit8.png


* This pattern has maxima where all cosine terms are either 1, or all are -1 .
o All cosines are +1, when π (d/λ) sinθ = 2 m π
o All cosines are -1, when π (d/λ) sinθ = (2 m + 1)π
* Hence, any integer number of π will do, and the directions of intensity maxima in pattern from diffraction grating are
sinθm = m λ/d
i.e. exactly the same as for two-slit interference.
* The value of m is called the order of the maximum .

This tells me that maxima can be found at angles of θm=sin-1(mλ/d) but it doesn't tell me anything about the number of fringes for an N = 2 interference pattern versus an N = 20 interference pattern.

Also, I assume there is a point where the bright fringes dissipate as the order of maximum gets larger and larger (m increases).
 
" and the directions of intensity maxima in pattern from diffraction grating are
sinθm = m λ/d
i.e. exactly the same as for two-slit interference. "

If the directions are the same as in case of two slits, the bright strips are at the same positions independent on the number of slits. If there are seven bright strips in between the minima of the diffraction pattern of a single slit, it will stay the same if you have a grating with many slits of the same width and spacing, only the bright strips will be narrower and brighter.

ehild
 
Thanks for the clarification ehild!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top