A wavelength of 625nm is used in a Young’s double-slit experiment. How many bright fringes lie on the screen?

Question

A wavelength of 625nm is used in a Young’s double-slit experiment. The separation between the slits is d = 1.4 × 10^−5 m. The total width of the screen is 0.20 m. In one version of the setup, the separation between the double slit and the screen is LA = 0.35m , whereas in another version it is LB = 0.50 m. On one side of the central bright fringe, how many bright fringes lie on the screen in the two versions of the setup? Do not include the central bright fringe in your counting.

So I'm not sure how to even attempt this problem. I thought about using the equation sin(theta)=mλ/d and just solving for m. I would get theta by using the distances to form a right triangle and take and inverse function of that but it ended in an error on my calculator. I'm really confused on this problem and I would really appreciate if someone could help me with the steps to manipulate the formula.

My professor's answers are

LA:22 and LB:16.

Answer 1

Hi Olivia!

 

I think you are on the right track.  See if the explanation below makes sense (even without a diagram), and see if it helps get you on track (at least to not getting a calculator error! :) )

 

The key here is that the angle θ that goes into the formula to find the location of maxima and minima is the same angle such that tan(θ) = y/L, where y is the straight-line distance from the center (bright spot) of the screen out to a given location along the screen, and L is the distance from the screen to the slits.

 

In this case, the maximum angle θ we can go off-axis and still hit the screen is given by:

 

tan(θ_max) = (half-width of screen)/L   (the half-width of the screen is the distance from the center to the edge)

 

so θ_max = tan^-1[(half-width of screen)/L] in each case.

 

This represents the maximum angle (from the horizontal) that light can leave the slits and still hit the screen.  If you put this into the interference maximum formula for double slits, you get:

 

dsin(θ_max) = xλ   (with d and λ given)

 

You can solve this for a value of x, which will probably not be an integer.  However, it does represent the dividing line between any bright fringes that hit the screen, and those that do not.  The largest integer (m) value less than x represents the number of the last bright fringe that can land on the screen.  It is therefore also the number of bright fringes between the center of the screen and the end (discounting the center -- which I often informally call the m=0 -- bright fringe, as the questions instructs).  Since this is symmetric on either side of center, double this number to get the total number of bright fringes on the screen.

 

I think this derivation is correct with the given information.  However, I did not quite get your instructor's answers when I did this.  I got 12 fringes total for L_A and 8 fringes total for L_B (not including the center fringe in either case).

 

I was able to get your instructor's answers if I used the full width of the screen (0.2 m) in the inverse tangent expression above, where I said to use half the width.  I think half the width is correct for the definition of the angles in this situation, since it corresponds to the angle away from the optical axis, which goes to the center of the screen.  So, if the given information is accurate, I think your instructor may have made small error.

 

With the given information, I think my results are correct.  But if I am mistaken, I apologize.

 

Just let me know if you have more questions about this!

 

Answer 2

I just want to add that I did the calculations too, using the same geometry that Steven explained.

d = slit separation = 1.4 x 10^-5 m

w = screen width = 0.20 m
L = slit to screen distance

 

For L = 0.35 m,

 

θ_max = tan^-1[w/(2L)] = tan^-1[0.20/(2*0.35)] = 15.945º

m ≤ (d/λ) sinθ_max = [(1.4 x 10^-5)/(625 x 10^-9)] sin15.945º = 6.15

 

Then m = 6, and we get 6 bright fringes on the half screen, hence 12 on the full screen

 

For L = 0.50 m,

 

θ_max = tan^-1[w/(2L)] = tan^-1[0.20/(2*0.50)] = 11.310º

 

m ≤ (d/λ) sinθ_max = [(1.4 x 10^-5)/(625 x 10^-9)] sin11.310º = 4.39

 

Then m = 4, and we get 4 fringes on the half screen, hence 8 on the full screen.

 

I verify Steven's results, and agree there is an error in the instructor's answers.