Physics, 20.04.2020 23:50 michellemonroe012305

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.
Assuming that the screen is sufficiently bright, at what distance can you no longer resolve two pixels on diagonally opposite corners of the screen, so that the entire screen looks like a single spot? Note that the size (0.360 meters) quoted for a monitor is the length of the diagonal.
Express your answer in meters to three significant figures.

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Answer from: JorgeW5St

Complete question:

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.

A) 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, 550*10^-9m as a characteristic optical wavelength. Express your answer in millimeters to three significant figures.

B.) Assuming that the screen is sufficiently bright, at what distance can you no longer resolve two pixels on diagonally opposite corners of the screen, so that the entire screen looks like a single spot? Note that the size (0.360 meters) quoted for a monitor is the length of the diagonal. Express your answer in meters to three significant figures.

a) 3.1mm

b) 1663m

Explanation:

Given:

Screen resolution =1024*768 pixel at approximately 281m each

a) Let's take sin∅ = $1.22\frac{wavelength}{d}$

$sin(tan^-^1(\frac{281*10^-^6}{1.30})= \frac{1.22*550*10^-^9}{d}$

Solving for d we have:

d = 3.10mm

b) given:

y = 0.360m

We now have d= 3.1mm

To find L, we use:

$sin(tan^-^1(frac{y}{L})$

$L = \frac{1.22*550*10^-^9}{3.1*10^-^3}$

L = 1663m

Answer from: Quest

hi there!

out of the two, (b) precision would be more reasonable. however, both a and b could be pretty reasonable.

-awesomerepublic : )

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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