Question

The pupil of the eye is the circular opening through which light enters. Its diameter can vary from about 8.00 mm to about 2.00 mm to control the intensity of the light reaching the interior. Calculate the angular resolution, ?R, of the eye for light that has a wavelength of 543 nm in both bright light and dim light.Dim light=______0 Bright light=__________0 At which light level is diffraction less of a limiting factor in the sharpness of your vision?A)BrightB) DimC) More information is needed to answerD) Diffraction is equally a factor in both intensities of light

Answer 1

Answer:

$0.0047^{\circ}$

$0.018978^{\circ}$

B) Dim

Explanation:

$\theta$ = Angle

a = Pupil diameter

$\lambda$ = Wavelength = 543 nm

Angular resolution is given by

$sin\theta=1.22\dfrac{\lambda}{a}\\\Rightarrow \theta=sin^{-1}\left(1.22\dfrac{\lambda}{a}\right)\\\Rightarrow \theta=sin^{-1}\left(1.22\times \dfrac{543\times 10^{-9}}{8\times 10^{-3}}\right)\\\Rightarrow \theta=0.0047^{\circ}$

For the dim light the angle is $0.0047^{\circ}$

$sin\theta=1.22\frac{\lambda}{a}\\\Rightarrow \theta=sin^{-1}\left(1.22\dfrac{\lambda}{a}\right)\\\Rightarrow \theta=sin^{-1}\left(1.22\times \dfrac{543\times 10^{-9}}{2\times 10^{-3}}\right)\\\Rightarrow \theta=0.018978^{\circ}$

For the bright light the angle is $0.018978^{\circ}$

The angular separation and sharpness are inversely related. Here, the dim light will produce a sharp image.

Hence, option B is correct