contestada

A space-based telescope can achieve a diffraction-limited angular resolution of 0.05″ for red light (wavelength 700 nm). What would the resolution of the instrument be (a) in the infrared, at 3.5 µm, and (b) in the ultraviolet, at 140 nm?

Respuesta :

Answer:

a) [tex] \theta_2 = 0.05 * \frac{3.5}{0.7} = 0.25[/tex]

b) [tex] \theta_2 = 0.05 * \frac{140}{700} = 0.01[/tex]

Explanation:

We are comparing two wavelengths with the radius and diameter constant, and if we want to compare it, we need to use the following formula:

[tex]\frac{\theta_1}{\theta_2}= \frac{\lambda_1}{\lambda_2}[/tex]

Where [tex] \theta[/tex] represent the angular resolution and [tex]\lambda[/tex] the wavelength.

So if we have a fixed resolution and wavelength 1 and we want to find the resolution for a new condition we can solve for [tex] \theta_2[/tex] and we got

[tex] \theta_2 = \theta_1 \frac{\lambda_2}{\lambda_1}[/tex]

Part a

For this case the subindex 1 is for the color red and we know that:

[tex] \lambda_1 = 700 nm *\frac{1 \mu m}{1000 nm} = 0.7 \mu m[/tex]

And the angular resolution for the color red is specified as [tex] \theta_1 = 0.05[/tex]

And for the infrared case we know that [tex] \lambda_2 = 3.5 \mu m[/tex], so if we replace we got:

[tex] \theta_2 = 0.05 * \frac{3.5}{0.7} = 0.25[/tex]

Part b

For this case the subindex 1 is for the color red and we know that:

[tex] \lambda_1 = 700 nm[/tex]

And the angular resolution for the color red is specified as [tex] \theta_1 = 0.05[/tex]

And for the ultraviolet case we know that [tex] \lambda_2 = 140 nm[/tex], so if we replace we got:

[tex] \theta_2 = 0.05 * \frac{140}{700} = 0.01[/tex]