To calculate the intensity at an angular position $\theta = 7.8^\circ$ for a single-slit diffraction pattern, we use the following formula for intensity:

$I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2$

where:

• $I_0$ is the maximum intensity,
• $\beta = \frac{\pi a}{\lambda} \sin(\theta)$,
• $a$ is the slit width (3.0 μm = $3.0 \times 10^{-6}$ m),
• $\lambda$ is the wavelength of light (589 nm = $589 \times 10^{-9}$ m),
• $\theta$ is the angle (7.8°).

Steps:

1. Convert angle to radians: $\theta = 7.8^\circ \times \frac{\pi}{180} = 0.136 \text{ radians}$

2. Calculate $\beta$: $\beta = \frac{\pi \times 3.0 \times 10^{-6}}{589 \times 10^{-9}} \times \sin(0.136) = \frac{9.42 \times 10^{-6}}{589 \times 10^{-9}} \times 0.136 \approx 0.022$

3. Now, use the formula for intensity: $I(\theta) = I_0 \left( \frac{\sin(0.022)}{0.022} \right)^2$

4. Since $\sin(0.022) \approx 0.022$, we have: $I(\theta) \approx I_0 \left( \frac{0.022}{0.022} \right)^2 = I_0$

Thus, the intensity at $\theta = 7.8^\circ$ is approximately the maximum intensity, $I_0$. It remains relatively unaffected at this angle for small values of $\beta$.

Would you like more details or have any questions on this?

Here are some related questions:

1. What happens to the diffraction pattern if the wavelength is increased?
2. How would the intensity change if the slit width was reduced?
3. How do you calculate the angular position of the first minimum in a single-slit diffraction pattern?
4. What is the significance of the first minimum in single-slit diffraction?
5. How does the intensity formula for single-slit diffraction differ from that of double-slit diffraction?

Tip: In diffraction problems, smaller slit widths or longer wavelengths tend to cause broader diffraction patterns.