Intensity in Single-Slit Diffraction, Quantitatively — Question 2

Question 2

The full width at half maximum is smaller than the first-zero width

For $I/I_c=(\sin\beta/\beta)^2$, the positive central half-maximum root is $\beta_h=1.391557378$. Use paraxial screen geometry $\beta=\pi ay/(\lambda L)$. Let $a=0.200\,\mathrm{mm}$, $L=2.00\,\mathrm m$, and $\lambda=600\,\mathrm{nm}$.

1. Derive and calculate the full width at half maximum.

2. Compare it with the width between the first zeros and give their ratio.

Question 2 - Solution

The full width at half maximum is smaller than the first-zero width

(a) Method and calculation

$$y_h=\frac{\beta_h\lambda L}{\pi a}\Longrightarrow\boxed{W_{1/2}=\frac{2\beta_h}{\pi}\frac{\lambda L}a}.$$$$\boxed{W_{1/2}=0.885893\frac{\lambda L}a=5.31536\,\mathrm{mm}}.$$

(b) Interpretation and further calculation

$$\boxed{W_{\mathrm{zero}}=2\lambda L/a=12.000\,\mathrm{mm}}.$$$$\boxed{W_{1/2}/W_{\mathrm{zero}}=\beta_h/\pi=0.442946}.$$

Physical check

Quoting a diffraction width without identifying its convention can introduce more than a factor-of-two discrepancy.