Shining a laser pointer towards the star vega

[moonkey]

Homework Statement

(1) A Physics lecturer, on a dark night, shines his laser pointer towards the star Vega.
(a) Roughly, how many optical photons from the Sun per second enter the eye of an
astronomer standing on a planet orbiting Vega ? Assume that 50 % of the light
from the Sun is generated in the optical passband and that our extra-terrestrial
astronomer has the same pupil diameter as that of a human.
(b) Roughly, how many photons are generated per second by the laser pointer ? The
pointer works in the wavelength range 620-680 nm, at a power of 5 mW, with an
aperture of 1 mm.
(c) If the light of the laser pointer is perfectly parallel, how many of these photons per
second will enter the eye of our extra-terrestrial astronomer ? Will he see them ?
(Assume that the extra-terrestrial astronomer can just see the Sun without a
telescope, and that the laser points directly at him).

Distance to Vega = 8.1 pc
Speed of light (c) = 3×108 m s−1
Planck constant (h) = 6.63 × 10−34 J s
Solar mass (M⊙) = 2 × 1030 kg
Solar Luminosity = 3.85 × 1026 W
1 pc = 3.086 × 1016 m

Homework Equations

The Attempt at a Solution

I have solved part (a) and (b). It's part (c) I'm having aproblem with.

Can I take it that the Luminosity of the laser is 5mW?

The light of the laser pointer is perfectly parallel so would the Luminosity to the observer be 5mW*πr²/8.1pc, where πr² is the area of the observer's eye. Like I said I solved part (b) so I know the number of photons per second. (I know I need to convert to meters first)?

 

[collinsmark]

I have solved part (a) and (b). It's part (c) I'm having aproblem with.

Can I take it that the Luminosity of the laser is 5mW?

I would think so, yes.

The light of the laser pointer is perfectly parallel so would the Luminosity to the observer be 5mW*πr²/8.1pc, where πr² is the area of the observer's eye. Like I said I solved part (b) so I know the number of photons per second. (I know I need to convert to meters first)?

I think you're supposed to assume that all of the energy of the laser arrives at Vega (technically the eye of the observer on a planet orbiting Vega) in a diameter of 1 mm. In other words, all of the power originating from the laser pointer is contained in a diameter of 1 mm when it leaves the laser; and it is still contained in a diameter of 1 mm when it arrives at its destination.

Above, you have the power from the laser pointer spreading out as a function from the radius (such that the power flux [through a given sized area] is a function of the distance). But I don't think you're supposed to do that. It would be okay for a isotropically radiating object such as the sun, but not for a directed source such as the laser. I interpret "the light of the laser pointer is perfectly parallel" as saying that the light (and its energy) doesn't spread out as it travels.

And regarding a different point, concerning something like the sun, which is radiating isotropically from its center (for all practical purposes anyway), the energy per unit area is proportional to inverse of the distance squared. 'Not simply \propto 1/d. It's \propto 1/d². This relationship doesn't apply to this ideal situation with the laser (since it is perfectly directed) but it would apply to the sun.

 

[moonkey]

Thank you very much for the reply. Helped a lot