The question in the image asks to calculate the smallest object that can be resolved on Earth by an optical sensor using the Rayleigh criterion.

Given dataThe question in the image asks to calculate the smallest object that can be resolved on Earth by an optical sensor using the Rayleigh criterion.

Given

• Wavelength ($\lambda$) = 500 nm = $500 \times 10^{-9}$ m
• Diameter of the optical aperture ($D$) = 4 m
• Height ($h$) = 800 km = $800 \times 10^3$ m

Formula:

Using the Rayleigh criterion, the angular resolution ($\alpha$) can be calculated as: $\sin(\alpha) = \frac{\lambda}{D}$ For small angles, $\sin(\alpha) \approx \alpha$, so: $\alpha = \frac{\lambda}{D}$

Next, to calculate the smallest object ($x$) that can be resolved on the ground, we use the following approximation: $x = h \cdot \alpha$

Now, I will calculate the result based on this information.The smallest object that can be resolved on Earth by this optical sensor is approximately 0.1 meters (10 cm).

Would you like more details on the calculation or further clarifications?

Here are 5 related questions to consider:

1. How does changing the diameter of the aperture affect the resolution?
2. How would the resolution change if the sensor were placed at a different height?
3. What happens to the resolution if the wavelength of the light used changes?
4. Can this method be used for non-optical wavelengths (e.g., radio waves)?
5. How is the Rayleigh criterion applied in astronomy?

Tip: In practical situations, atmospheric distortion (known as "seeing") can further limit the resolution of optical systems from Earth.