# Diffraction Problem

1. Dec 7, 2003

Tried this problem for about 2 hrs and no dice...

A spy satellite orbiting at 150 km above the Earth's surface has a lens with a focal length of 3.5 m and can resolve objects on the ground as small as 36 cm; it can easily measure the size of an aircraft's air intake. What is the effective lens diameter, determined by diffraction consideration alone? Assume = 527 nm.

I dont think the 150km matters...
i just did .36 = f*theta and i got theta = .36/3.5

then i did theta = wavelength /d (diameter) and i get like 5.12e-6M

totally wrong...help

btw as soon as someone replies i will prolly reply back since its kinda urgent

Last edited: Dec 7, 2003
2. Dec 8, 2003

### chroot

Staff Emeritus
Consider the Rayleigh criterion for a circular aperture:

$$\sin \theta_R = 1.22 \frac{\lambda}{d}$$

In this case, $\theta_R$ is the angle subtended by a 36 cm feature from 150 km away. Hint: draw a right triangle. The angle subtended is

\begin{align*} \theta_R &= 2 \tan^{-1} \left( \frac{0.18}{150 \cdot 10^3} \right)\\ & \approx \tan^{-1} \left( \frac{0.36}{150 \cdot 10^3} \right)\\ & \approx 2.4 \cdot 10^{-6}\ \text{rad} \end{align*}

The Rayleigh criterion then gives

\begin{align*} d &= \frac{1.22 \lambda}{\sin \theta_R}\\ & \approx 0.27\ m \end{align*}

Does this make sense?

- Warren