Question

A standard 14.16-inch (0.360-meter) computer monitor is 1024 pixels wide and 768 pixels tall. Each pixel is a square approximately 281 micrometers on each side. Up close, you can see the individual pixels, but from a distance they appear to blend together and form the image on the screen.

If the maximum distance between the screen and your eyes at which you can just barely resolve two adjacent pixels is 1.30 meters, what is the effective diameter d of your pupil? Assume that the resolvability is diffraction-limited. Furthermore, use lambda = 550 nanometers as a characteristic optical wavelength.

Answer 1

Answer:

0.0031 m

Explanation:

y = Length of pixel = 281 μm

L = Distance to screen = 1.3 m

$\lambda$ = Wavelength = 550 nm

d = Pupil diameter

$\theta$ = Angle

We have the expression

$tan\theta=\dfrac{y}{L}\\\Rightarrow \theta=tan^{-1}\dfrac{y}{L}\\\Rightarrow \theta=tan^{-1}\dfrac{281\times 10^{-6}}{1.3}$

We have the expression

$sin\theta=1.22\dfrac{\lambda}{d}\\\Rightarrow d=\dfrac{1.22\lambda}{sin\theta}\\\Rightarrow d=\dfrac{1.22\times 550\times 10^{-9}}{sin\left(tan^{-1}\frac{281\times 10^{-6}}{1.3}\right)}\\\Rightarrow d=0.0031\ m$

The pupil diameter is 0.0031 m

Answer 2

The  effective diameter d of your pupil should be considered as the 0.0031 when maximum distance should be 1.30 meters.

Calculation of the effective diameter:

Since

y = Length of pixel = 281 μm

L = Distance to screen = 1.3 m

Wavelength = 550 nm

d = Pupil diameter

So,

$tan = y / l\\\\= tan^-1 * 281*10^-6/1.3$

Now

sin = 1.22 wavelength / d

d = 1.22 * wavelength / sin

$= 1.22*550*10^-9/ sin ( tan^-1 * 281*10^-6/1.3)$

= 0.0031 m

hence, The  effective diameter d of your pupil should be considered as the 0.0031 when maximum distance should be 1.30 meters.

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