Question

The threshold intensity for a human eye to sense the presence of light is 50 photons falling on the retina within a period of 0.1 s. Assume a star radiates with an output of 3 ×1026 W in the visible spectrum, with a peak of λ=550 nm. (This star would be similar to the Sun, which radiates with 2.0 ×1026 W). Estimate how far away the star can be away and still be visible to the unaided eye on a dark night. Assume the pupil of the observer's eye has a diameter of 1.0 cm. Give your answer in light years (do not insert a unit).

Answer 1

Answer 2

To estimate how far away the star can be and still be visible to the unaided human eye on a dark night, we need to consider the intensity of light reaching the observer's eye from the star.

First, we can calculate the intensity of light (I) at the observer's eye using the radiant flux (power) of the star and the surface area of a sphere at a given distance (r) from the star.

The intensity (I) of light at a distance (r) from the star is given by the formula:

I = L / (4πr^2)

Where: L = Radiant flux (power) of the star (in watts) r = Distance from the star (in meters)

Given: Radiant flux of the star (L) = 3 × 10^26 W Peak wavelength (λ) = 550 nm = 550 × 10^-9 m (in meters)

Next, we can calculate the number of photons reaching the observer's eye in 0.1 seconds (Δt) based on the threshold intensity of the human eye.

Number of photons (N) = Threshold intensity × Area of the pupil × Δt

Given: Threshold intensity = 50 photons / 0.1 s = 500 photons/s Area of the pupil = π × (0.5 cm)^2 = π × (0.5 × 10^-2 m)^2

Now, we can equate the number of photons to the intensity and solve for the distance (r):

N = I × Area of the pupil × Δt

Substitute the values:

500 photons/s = (L / (4πr^2)) × (π × (0.5 × 10^-2 m)^2) × 0.1 s

Now, solve for the distance (r):

r^2 = L / (500 × 0.1 × (0.5 × 10^-2)^2)

r^2 = (3 × 10^26) / (500 × 0.1 × (0.5 × 10^-2)^2)

r^2 ≈ 1.2 × 10^29

r ≈ √(1.2 × 10^29) ≈ 1.1 × 10^15 meters

Finally, convert the distance from meters to light years:

1 light year ≈ 9.461 × 10^15 meters

Distance ≈ (1.1 × 10^15) / (9.461 × 10^15) ≈ 0.116 light years

Therefore, the star can be approximately 0.116 light years away and still be visible to the unaided human eye on a dark night.