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Wave Optics Interference Experiment

Created by @nathanedwards
 at November 3rd 2023, 9:51:01 am.
AP_Physics_2-Questions
#wave-optics
#interference
#double-slit-experiment

AP Physics 2 Exam Question - Wave Optics

Question:

A student is performing an experiment to investigate the interference of light using a double-slit setup. The apparatus consists of two narrow slits separated by a distance of d=1.0 mmd = 1.0 \, \text{mm}. The student sets up a screen at a distance of L=1.5 mL = 1.5 \, \text{m} from the double slit and observes the interference pattern of bright and dark fringes on the screen.

a) Calculate the spacing between adjacent bright fringes on the screen.

b) If the wavelength of the light used in the experiment is λ=600 nm\lambda = 600 \, \text{nm}, calculate the angular position at which the third-order bright fringe occurs.

c) The student decides to increase the separation between the slits to d′=2.0 mmd' = 2.0 \, \text{mm}. Determine how this change affects the interference pattern on the screen. Provide a qualitative explanation.

Answer:

a) The spacing between adjacent bright fringes can be calculated using the formula:

Spacing between adjacent bright fringes=λ⋅Ld \text{Spacing between adjacent bright fringes} = \frac{\lambda \cdot L}{d}

where λ\lambda is the wavelength of light, LL is the distance from the double slit to the screen, and dd is the separation between the slits.

Substituting the given values:

Spacing between adjacent bright fringes=(600×10−9 m)⋅(1.5 m)1.0×10−3 m=0.009 m or 9.0 mm \text{Spacing between adjacent bright fringes} = \frac{(600 \times 10^{-9} \, \text{m}) \cdot (1.5 \, \text{m})}{1.0 \times 10^{-3} \, \text{m}} = 0.009 \, \text{m} \, \text{or} \, 9.0 \, \text{mm}

Therefore, the spacing between adjacent bright fringes on the screen is 9.0 mm9.0 \, \text{mm}.

b) The angular position at which the mm-th order bright fringe occurs can be calculated using the formula:

θ=atan(m⋅λd) \theta = \text{atan}\left(\frac{m \cdot \lambda}{d}\right)

where λ\lambda is the wavelength of light, dd is the separation between the slits, and mm is the order of the bright fringe.

Substituting the given values (m=3m = 3):

θ=atan(3⋅(600×10−9 m)1.0×10−3 m) \theta = \text{atan}\left(\frac{3 \cdot (600 \times 10^{-9} \, \text{m})}{1.0 \times 10^{-3} \, \text{m}}\right)

Using a calculator, we find:

θ≈1.732 degrees \theta \approx 1.732 \, \text{degrees}

Therefore, the angular position at which the third-order bright fringe occurs is approximately 1.7321.732 degrees.

c) Increasing the separation between the slits to d′=2.0 mmd' = 2.0 \, \text{mm} will decrease the spacing between adjacent bright fringes on the screen according to the formula:

Spacing between adjacent bright fringes=λ⋅Ld′ \text{Spacing between adjacent bright fringes} = \frac{\lambda \cdot L}{d'}

As d′d' increases, the denominator in the formula increases, resulting in a smaller spacing between adjacent bright fringes. This means that the fringes will appear closer together on the screen.

Additionally, increasing the separation between the slits will increase the number of bright fringes that can be observed on the screen. The overall pattern will have more bright fringes, but the distance between them will be smaller compared to the initial setup.

This change also leads to a wider central maximum, as the width of the central maximum is determined by the overall spread of the bright fringes.

In summary, increasing the separation between the slits will result in a closer spacing between the bright fringes on the screen, a higher number of bright fringes, and a wider central maximum.

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