Diffraction — Question 9

Question 9

A lens makes angular diffraction accessible on a nearby plane

A thin converging lens is placed immediately after a slit. Parallel rays leaving the slit at small angle $\theta$ meet at $y=f\theta$ in the rear focal plane. The slit has $a=0.25\,\mathrm{mm}$, wavelength $625\,\mathrm{nm}$, and the lens has $f=0.40\,\mathrm m$. Assume a sufficiently large lens aperture.

1. Find the focal-plane position of the first slit minimum.

2. If the detector pixels are $10\,\mu\mathrm m$ wide, calculate the number of pixels across the central maximum and how it changes for a lens with twice f.

Question 9 - Solution

A lens makes angular diffraction accessible on a nearby plane

(a) Method and calculation

$$a\theta_1\simeq\lambda\Longrightarrow y_1=f\lambda/a.$$$$\boxed{y_1=1.00\,\mathrm{mm},\quad W=2y_1=2.00\,\mathrm{mm}}.$$

(b) Interpretation and further calculation

$$\boxed{N=W/p=2.00\,\mathrm{mm}/(0.010\,\mathrm{mm})=200}.$$$$f\to2f\Longrightarrow\boxed{W\to4.00\,\mathrm{mm},\quad N\to400}.$$

Physical check

The lens maps direction into position; a physical screen at one focal length can display the angular spectrum without a very long free-space propagation distance.