Diffraction lens diameter Problem

[purduegrad]

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 don't 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

 

[chroot]

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

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

The Rayleigh criterion then gives

$$\begin{align*}<br /> d &= \frac{1.22 \lambda}{\sin \theta_R}\\<br /> & \approx 0.27\ m<br /> \end{align*}$$

Does this make sense?

- Warren

 

[M98Ranger]

First of all, I would like to commend you for putting in the effort to solve this problem for 2 hours. Diffraction problems can be tricky, so don't get discouraged if you haven't found the correct answer yet.

To solve this problem, we need to use the diffraction limit formula: θ = 1.22λ/D, where θ is the angular resolution, λ is the wavelength, and D is the diameter of the lens.

Since we are given the focal length of the lens (3.5 m) and the smallest resolvable object (36 cm), we can use the formula θ = 1.22λ/f to find the angular resolution. Plugging in the values, we get:

θ = 1.22 * 527 nm / 3.5 m = 1.83e-4 radians

Now, we can use the formula θ = λ/D to find the effective lens diameter. Rearranging the formula, we get:

D = λ/θ = 527 nm / 1.83e-4 = 2.88 m

Therefore, the effective lens diameter, determined by diffraction consideration alone, is approximately 2.88 meters. This is larger than the actual lens diameter of 3.5 m, which means that diffraction is not the only factor limiting the resolution of the spy satellite.

I hope this helps you understand the problem better. Keep practicing and don't give up, you will eventually get the hang of diffraction problems!