Diffraction — Question 8

Question 8

Narrowing an illuminated slit trades peak brightness for angular spread

In a one-dimensional scalar far-field model with fixed incident irradiance and fixed slit height, the on-axis amplitude is proportional to a, transmitted power to a, and central angular width to $1/a$. Reduce the slit width from a to $a/4$.

1. Calculate the ratios of peak intensity, central angular width, and transmitted power.

2. Use the rescaled intensity $I_a(q)=C a^2F(aq)$, with fixed integrable shape F, to verify the total-power scaling by integration over q.

Question 8 - Solution

Narrowing an illuminated slit trades peak brightness for angular spread

(a) Method and calculation

$$E_{\mathrm{peak}}^\prime/E_{\mathrm{peak}}=1/4\Longrightarrow\boxed{I_{\mathrm{peak}}^\prime/I_{\mathrm{peak}}=1/16}.$$$$\boxed{\Delta\theta^\prime/\Delta\theta=4,\quad P^\prime/P=1/4}.$$

(b) Interpretation and further calculation

$$\int_{-\infty}^{\infty}I_a(q)dq=Ca^2\int F(aq)dq=Ca\int F(v)dv.$$$$\boxed{P\propto a,\quad P^\prime/P=(a/4)/a=1/4}.$$

Physical check

The reduced peak is partly offset by spreading over more angles; multiplying peak scaling by width scaling gives the correct power ratio.