Single Slit Diffraction Patterns

A red laser with $\lambda = 650\text{ nm}$ passes through a slit of width $a = 0.1\text{ mm}$. Find the angle $\theta$ of the first minimum.

1. Convert units to meters: $\lambda = 6.5 \times 10^{-7}\text{ m}$ and $a = 1.0 \times 10^{-4}\text{ m}$. 2. Use the formula for $n=1$: $a \sin \theta = (1)\lambda$. 3. Solve for $\sin \theta$: $\sin \theta = \frac{6.5 \times 10^{-7}}{1.0 \times 10^{-4}} = 0.0065$. 4. Calculate $\theta$: $\theta = \arcsin(0.0065) \approx 0.37^\circ$.

If the screen is at a distance $D = 2.0\text{ m}$ from the slit, what is the physical width ($w$) of the central maximum for the laser in the previous example?

1. Recall the angle to the first minimum: $\theta_1 \approx 0.0065\text{ radians}$. 2. The distance from the center to the first minimum is $y_1 = D \tan \theta_1 \approx D\theta_1$. 3. $y_1 = 2.0 \times 0.0065 = 0.013\text{ m} = 1.3\text{ cm}$. 4. The total width $w$ is $2y_1 = 2 \times 1.3 = 2.6\text{ cm}$.

A student observes a diffraction pattern. If they double the slit width $a$, how do the intensity and the width of the central maximum change?

1. Width: Since $\theta \approx \frac{\lambda}{a}$, doubling $a$ halves the angle $\theta$. The central maximum becomes twice as narrow. 2. Intensity: More light passes through a wider slit, and that light is concentrated into a smaller area. The peak intensity actually increases by a factor of 4 (Intensity $\propto a^2$).