Question

1. What is the minimum telescope aperture needed to resolve an object whose angular diameter is 0.38 arc seconds, observed at 540 nm wavelength? Note: 1arcsec=(1/3600)∘. **express in cm

2. Find the minimum angular separation resolvable with 625 nm laser light passing through a circular aperture of diameter 2.1 cm .

3. The interference pattern from two slits separated by 0.35 mm has bright fringes with angular spacing 0.073 ∘. **express in nm**

Answer 1

1.

Qmin = 1.22 l/D where Q is the minimum resolvable angel in radians, D = aperture diameter, and l = wavelength.

Convert arsec to radians ---> Qmin = 1.84224x10^-6 radians

Solve for D:

D = 1.22 l/Qmin = 1.22* (540x10^-9 m)/(1.84224x10^-6 rad) = 0.35760 m

2.

     θ = 1.220 λ / D                     

θ is the angular resolution (radians),
λ is the wavelength of light,
and D is the diameter of the lens' aperture

θ = 1.220 *625*10^-9 / 2.1*10^-2

θ = 3.630*10^-5⁰

3.

θ = λ/d
Convert mm to nm 10^-3

λ = Sin( .073) * 0.35 * 10^-3

    = 233 nm