An opaque sheet has a hole in it of $0.5 \mathrm{~mm}$ radius. If plane waves of light $(\lambda=5000 \AA)$ fall on the sheet, find the maximum distance from this sheet at which a screen must be placed so that the light will be focused to a bright spot.
The maximum distance $r$ of the screen from the hole for which the light will be focused to a bright spot is that for which the area of the hole corresponds to the first Fresnel zone only, and is given by

$$
\rho_{1}^{2}=\lambda r,
$$

where $\rho$ is the radius of the hole. Hence

$$
r=\frac{\rho_{1}^{2}}{\lambda}=\frac{0.5^{2}}{5000 \times 10^{-7}}=500 \mathrm{~mm} .
$$

Without the opaque screen, the number of the Fresnel zones will approach infinity. The total amplitude is then

hence

$$
\begin{aligned}
A &=A_{1}-A_{2}+A_{3}-\ldots \ldots \\
&=\frac{A_{1}}{2}+\left(\frac{A_{1}}{2}-A_{2}+\frac{A_{3}}{2}\right)+\left(\frac{A_{3}}{2}-A_{4}+\frac{A_{5}}{2}\right)+\ldots \\
& \approx \frac{A_{1}}{2}
\end{aligned}
$$

$$
I^{\prime} \sim A_{1}^{2} / 4 \text {, i.e., } I^{\prime}=I_{1} / 4
$$

Thus the intensity is now $1 / 4$ that before the opaque sheet is removed.