Shining a laser pointer towards the star vega

• moonkey
In summary, a Physics lecturer is shining a laser pointer towards the star Vega on a dark night. The conversation discusses the number of optical photons from the sun and the laser pointer that enter the eye of an astronomer standing on a planet orbiting Vega. The distance to Vega, the speed of light, Planck constant, solar mass, solar luminosity, and the conversion of pc to meters are mentioned. The expert suggests that the power from the laser pointer does not spread out as it travels and that the energy per unit area is proportional to the inverse of the distance squared for isotropically radiating objects such as the sun.
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

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)?

moonkey said:
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.

Thank you very much for the reply. Helped alot

What is the purpose of shining a laser pointer towards the star Vega?

Shining a laser pointer towards the star Vega allows for the measurement of atmospheric turbulence and the calculation of the star's distance from Earth. It is also used to demonstrate the phenomenon of light scattering.

How is a laser pointer able to reach a star like Vega?

A laser pointer is able to reach a star like Vega because it emits a highly concentrated beam of light that is able to travel long distances. Additionally, Vega is one of the brightest stars in the night sky, making it easier for the laser beam to be seen from Earth.

What type of laser is used to shine towards Vega?

The type of laser used to shine towards Vega is typically a green laser, as our eyes are more sensitive to green light and it is easier to see the beam. However, any color laser can be used as long as it is powerful enough to be seen from Earth.

Are there any potential risks involved in shining a laser pointer towards Vega?

Yes, there are potential risks involved in shining a laser pointer towards Vega. The laser beam could potentially interfere with aircraft navigation or other astronomical observations. It is important to always use caution and follow safety guidelines when shining a laser towards any celestial object.

Can shining a laser pointer towards Vega provide any valuable scientific data?

Yes, shining a laser pointer towards Vega can provide valuable scientific data. The measurement of atmospheric turbulence can be used to study the Earth's atmosphere, and the calculation of Vega's distance can contribute to our understanding of the universe and its expansion.

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