Finding the Aperture Size for 1-Arcsec Resolution at 500nm Wavelength

[Benzoate]

Homework Statement

What apeture is required to give 1-arcsec resolution for a wavelength of 500 nm(visible)

Homework Equations

theta=wavelength/d ; d is the size of the apeture

The Attempt at a Solution

I'm assuming when they are asking for the apeture , they are asking for the size of the apeture.

theta=wavelength/d => d=wavelength/theta =

I converted 1 arc second to 5*10^-6 radian

d=500*10^-9 m/5*10^-6= .1 m

 

[learningphysics]

Looks right to me.