Here is a solution and explanation for the activity questions based on the provided image:

1. Determine the width of the slits and the distance between the slits.

The double-slit intensity pattern is described by:

$I(\theta) = I_0 \left(\cos^2\left(\frac{\pi d \sin\theta}{\lambda}\right)\right) \cdot \left(\frac{\sin^2\left(\frac{\pi a \sin\theta}{\lambda}\right)}{\left(\frac{\pi a \sin\theta}{\lambda}\right)^2}\right)$

Where:

• $d$: Distance between the slits.
• $a$: Width of the slits.
• $\lambda = 633 \, \text{nm}$ (wavelength of light used).
• $\theta$: Angular position.

From the graph:

• The maxima spacing ($\Delta\theta$) in the double-slit interference pattern corresponds to the distance between maxima. Here, maxima are spaced at approximately $0.01 \, \text{radians}$, indicating the value of $d$ using the condition for constructive interference: d \sin \theta = m\lambda \quad \text{(where m is the order of the maxima)}.

For small angles ($\sin\theta \approx \theta$): $d = \frac{\lambda}{\Delta\theta} = \frac{633 \times 10^{-9}}{0.01} \approx 63.3 \, \mu\text{m}.$

• The minima spacing corresponds to the diffraction envelope caused by slit width ($a$). The first diffraction minima occurs at: $a \sin\theta = m\lambda \quad (m = 1).$

Using the location of the first minima (approximately $0.005 \, \text{radians}$): $a = \frac{\lambda}{\theta} = \frac{633 \times 10^{-9}}{0.005} \approx 126.6 \, \mu\text{m}.$

Answer:

• Width of slits ($a$): $126.6 \, \mu\text{m}$.
• Distance between slits ($d$): $63.3 \, \mu\text{m}$.

2. Flaw in the student's reasoning.

The student's comment:

• "To determine slit width, I used the first minimum at 0.005 radians, and to determine the distance between the slits, I used the first maximum at 0.01 radians."

Flaw: The student is conflating the two phenomena:

1. The minima at $0.005 \, \text{radians}$ correspond to the diffraction envelope created by the slit width ($a$).
2. The maxima at $0.01 \, \text{radians}$ are due to the interference pattern created by the distance between the slits ($d$).

These are separate effects governed by different equations. Using maxima to determine slit width or minima to determine slit separation is incorrect.

3. Accounting for the "missing" maxima.

Explanation: The "missing" maxima at $0.05 \, \text{radians}$ and $0.10 \, \text{radians}$ result from the diffraction envelope imposed by the slit width ($a$). The intensity pattern is modulated by a sinc-squared function: $\frac{\sin^2\left(\frac{\pi a \sin\theta}{\lambda}\right)}{\left(\frac{\pi a \sin\theta}{\lambda}\right)^2}.$

The missing maxima correspond to points where the interference maxima fall outside the main diffraction envelope (i.e., their intensity is zero due to the diffraction minima). If the slit width ($a$) were decreased, the diffraction envelope would broaden, allowing more maxima to appear.

Follow-up Questions:

1. Why does the interference pattern depend on both slit width and slit separation?
2. How does decreasing the wavelength of the light change the observed pattern?
3. What is the significance of the diffraction envelope in the intensity graph?
4. How can experimental errors affect the determination of $d$ and $a$?
5. What would happen to the pattern if the slits were moved farther apart?

Tip: Always separate the effects of interference (related to $d$) and diffraction (related to $a$) when analyzing a double-slit experiment.