The document describes a girl recreating Young's double slit experiment and observing 7 fringes with a certain slit spacing and deflection angle. It then provides answers to questions about adjusting the experiment to produce different numbers of fringes by changing the slit spacing or light wavelength. The answers explain that changing the slit spacing by a factor of 5/3 would produce 11 fringes, changing the wavelength by a factor of 3/2 would produce 5 fringes, and that using particles instead of light waves would not produce interference fringes on the screen.

1. A girl has re-created Young’s double slit experiment. At a certain slit spacing and deflection
angle, she observes 7 fringes on her screen.
For questions 1 and 2, always refer back to the original conditions (especially spectral
order) that produces 7 fringes.
1) If she wanted to see 11 fringes, by what factor could she change the original slit
spacing, d? Assume she keeps all other factors constant.
2) If she wanted to see 5 fringes, by what factor should she change the original light
wavelength? Assume she keeps all other factors constant.
3) If L and y are equal in length, and the slit spacing, d, is 5 m, how much longer is D1
than D2 ?
4) Instead of light waves, can a beam of small particles be used in Young’s double slit
experiment?

2. Answers:
1) The original value for the spectral order is 3, since the number of fringes is 7. In
other words, the 7 fringes can be labelled m= -3, -2, -1, 0, 1, 2, and 3.
To see 11 fringes, the value of m must be 5. This is achieved by multiplying our
original m by a factor of 5/3. Use the following equation:
dsinθ = mλ
Since we are told we can only manipulate d in this question, multiply the slit spacing
also by 5/3 (since the m value is directly proportional to d).
2) The original value for the spectral order is 3. To see 5 fringes, we obtain an m value
of 2 by multiplying m by 2/3. Since we are told we can only manipulate the light
wavelength, we multiply the wavelength by 3/2 (since the m value is inversely
proportional to the wavelength).
3) Since L and y are equal, then tanθ = y/L = 1. By taking the inverse tangent, θ = 45°.
The difference between D1 and D2 is equal to dsinθ. Since sinθ = sin45° ≃ 0.707 and
d is equal to 5 m, the difference between D1 and D2 ≃ 3.54 m.
4) Light can be used in this model since it has wave properties. Discovering this fact
was the purpose of Young’s experiment! A beam of particles would not create
fringes on the screen. Only particles that moved directly through the slits would be
able to pass through and hit the screen. The other particles would rebound off of the
slit plate. No wave fringe patterns would be observed.
Photo credit to: http://astro.physics.sc.edu/~rjones/phys153/lec06.html