Finding Maximum Number of Interference Maxima?

AI Thread Summary
The discussion revolves around calculating the maximum number of interference maxima observable when light of wavelength 535 nm passes through two slits 670 μm apart. The relevant equation is dsinθ = mλ, where the maximum angle is critical for determining the integer value of m. The calculated maximum angle is approximately 52.98°, leading to m being 1.26, which must be rounded down to 1 for integer values. Therefore, considering both sides of the central maximum, the total number of observable maxima is three: one central maximum and one on each side.
hrf2
Messages
5
Reaction score
0

Homework Statement


Light of wavelength λ = 535 nm shines through two narrow slits which are 670 μm apart. What is the maximum number of interference maxima which could conceivably be observed (assuming that diffraction minima do not extinguish them and the screen is arbitrarily large)?Your answer should be an integer. There is no sig-fig requirement for your answer

  1. Hint: What is the maximum angle allowed? Did you remember to count the maxima below the center?

Homework Equations


When I tried to answer the first time, the question gave me the a hint to look for the max angle allowed, so I think I'll need to use
dsinθ = mλ, where d=slit distance = 670 μm = 6.7 x 10-7m and λ=535nm = 5.35 x 10-7m
I also think I'll need y=mdL/λ, where y=distance between maxima, m= maxima integer, d=slit distance, L=distance between slits and screen (not given) and λ= wavelength, so maybe not this equation (because L isn't given?)
I thought about using wsinθ=mλ (where w=slit width), but because the question states to assume diffraction minima do not extinguish the maxima, I didn't think it was necessary to factor this in as well.

The Attempt at a Solution


Honestly I'm not even sure where to begin with this. I tried solving for the maximum angle as the question gave me feedback, and I got 52.98°. But I don't know if this is even relevant or what. For an earlier question I did m= d/λ and got the correct answer, but for some reason this does not work here.

Thanks in advance for any help!
 
Physics news on Phys.org
What can be the diffraction angle maximum ? Can it exceed 90°?
 
  • Like
Likes hrf2
If it's 90°, wouldn't the equation be dsin(90°) = mλ? And sin(90°) = 1
Rearranging for m: m=d/λ= (6.7 x 10-7m)/(5.35 x 10-7m) = 1.26. Is this the number of maxima produced? Or just for one side of the central maxima?
So I multiplied by 2 (for either side) and got 2, but that's wrong. So I'm getting confused somewhere.
 
hrf2 said:
If it's 90°, wouldn't the equation be dsin(90°) = mλ? And sin(90°) = 1
Rearranging for m: m=d/λ= (6.7 x 10-7m)/(5.35 x 10-7m) = 1.26. Is this the number of maxima produced? Or just for one side of the central maxima?
So I multiplied by 2 (for either side) and got 2, but that's wrong. So I'm getting confused somewhere.
Yes, m would be 1,26, but it must be integer. You need to tale the integer part. It is 1, so the orders of ±1 are observed, one maximum at both sides of the central maximum. The question was What is the maximum number of interference maxima which could conceivably be observed? . You observe the central maximum, too.
 
Last edited:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top