Question

Assume your eye has an aperture diameter of 3.00 mm at night when bright headlights are pointed at it. 1) At what distance can you see two headlights separated by 1.80 m as distinct? Assume a wavelength of 550 nm, near the middle of the visible spectrum. (Express your answer to two significant figures.)

Answer 1

The distance is 6259.31 meters.

Step-by-step explanation:

We shall use the Reyligh criterion to solve the problem

For diffraction due to circular aperture we have

Assuming that human eye is circular we have

`(x)/(D)=(1.22\lambda )/(d)`

`\therefore D=(xd)/(1.22\lambda )`

Applying the given values we have

`D=(1.40* 3* 10^(-3))/(1.22* 550* 10^(-9))\\\\\therefore D=6259.31m\\D=6.26km`

Answer 2

8.1 x 10³ m

Step-by-step explanation:

D = diameter of the aperture =3 mm = 0.003 m

x = distance between the two headlights = 1.80 m

λ = wavelength of the visible spectrum = 550 x 10⁻⁹ m

d = distance of the headlights from eyes = ?

Using Reyleigh's criterion,

`(x)/(d)=(1.22\lambda )/(D)`

`(1.80)/(d)=(1.22 (550* 10^(-9)) )/(0.003)`

d = 8047.7 m

d = 8.1 x 10³ m