# The pupil of an eagle's eye has a diameter of 6.0 mm. Two field mice are separated by 0.010 m....

## Question:

The pupil of an eagle's eye has a diameter of 6.0 mm. Two field mice are separated by 0.010 m. From a distance of 166 m, the eagle sees them as one unresolved object and dives toward them at a speed of 24 m/s. Assume that the eagle's eye detects light that has a wavelength of 550 nm in vacuum.

How much time passes until the eagle sees the mice as separate objects? Please give answer and show all steps.

## Resolving Power:

For any optical device, there is some resolution limit set by diffraction at the edges of the optical elements. The human, or eagle, eye is no different. The angular resolution {eq}\theta, {/eq} is given by {eq}\theta = \frac{1.22 \lambda}{D}. {/eq}

It depends on the particular wavelength {eq}\lambda {/eq} of light, as well as the size of the optical element {eq}D {/eq}.~

## Answer and Explanation:

Let's first look at the optics of the eagle. The resolution limit of the eagle's eyes, with diameter of 6.0 mm and seeing light of wavelength 550 nm, is

{eq}\theta = \frac{1.22 \lambda}{D} = \frac{1.22 \cdot 550 \times 10^{-9}~\textrm{m}}{6.0\times 10^{-3}~\textrm{m}} = 1.1 \times 10^{-4} {/eq} rad.

If the two mice are separated by a distance {eq}d = 0.010~\textrm{m}, {/eq} then the distance {eq}L {/eq} away the eagle must be to see them individually is

{eq}\tan \theta = \frac{d}{L} \rightarrow L = \frac{d}{\tan \theta} = \frac{0.010~\textrm{m}}{\tan(1.1 \times 10^{-4} \text{ rad})} = 90.9~\textrm{m}. {/eq}

So if the eagle is at present 166 metres away, he must swoop down to 90.9 metres away, in order to see his buffet.

The distance travelled by an object moving with velocity {eq}v {/eq} is {eq}x(t) = x_0 +v t {/eq}. If we choose a co-ordinate set at the mice and direct it toward the eagle, then {eq}x_0 = 166~\textrm{m}, {/eq} {eq}x(t) = 90.0~\textrm{m} {/eq} and {eq}v = -24~\textrm{m}~\textrm{s}^{-1} {/eq}.

Therefore the elapsed time is

{eq}x(t) = x_0 -vt \rightarrow t = \frac{x_0 - x(t)}{v} = \frac{166~\textrm{m} - 90.9~\textrm{m}}{24~\textrm{m}~\textrm{s}^{-1}} = 3.1~\textrm{s}. {/eq}