Diffraction — Question 1

Question 1

A slit width sets the angular scale of wave spreading

For a uniformly illuminated slit of width a, the first far-field zero obeys $a\sin\theta_1=\lambda$. A screen is distance $L=2.0\,\mathrm m$ away. Take $a=0.20\,\mathrm{mm}$ and $\lambda=500\,\mathrm{nm}$ in air.

1. Find the exact angle and screen distance of the first zero from the axis.

2. Find the full central width between the two first zeros and its change when a is halved, using the small-angle approximation.

Question 1 - Solution

A slit width sets the angular scale of wave spreading

(a) Method and calculation

$$\boxed{\theta_1=\sin^{-1}(\lambda/a)=\sin^{-1}(0.0025)=0.14324^\circ}.$$$$\boxed{y_1=L\tan\theta_1=5.000016\,\mathrm{mm}}.$$

(b) Interpretation and further calculation

$$W=2L\tan\theta_1\simeq\boxed{2L\lambda/a=10.0\,\mathrm{mm}}.$$$$a\to a/2\Longrightarrow\boxed{W\to2W=20.0\,\mathrm{mm}}.$$

Physical check

A narrower opening spreads the transmitted field through a larger angular range; the screen width is not the physical opening width.