Question

A person's near point is 25 cm. The person examines a mobile device monitor located at the near point. If the person's pupil is 5 mm in diameter, what is the minimum pixel separation of the monitor that the person can appreciate? Assume that the monitor illuminates green light at 550 nm. If the person looks at the mobile device placed in water, determine its minimum pixel separation at the near point. The refractive index of water is 1.33

Answer 1

minimum pixel separation of the monitor that the person is 0.3355 mm

minimum pixel separation at the near point in water is 0.2523 mm

Step-by-step explanation:

Given data

point u = 25 cm

diameter D = 5 mm

wavelength λ = 550 nm

refractive index of water μ = 1.33

to find out

what is the minimum pixel separation and minimum pixel separation at the near point

solution

we know that here

minimum pixel separation that is = 1.22 wavelength / μsin(θ)

minimum pixel separation = 1.22(550 ×
`10^(-9)` / μ(D/u)

minimum pixel separation = 1.22 (550 ×
`10^(-9)`) 25 ×
`10^(-2)` / 1 (5) ×
`10^(-3)`

and here μ = 1

minimum pixel separation = 33.55 ×
`10^(-6)` m

minimum pixel separation = 0.3355 mm

minimum pixel separation of the monitor that the person is 0.3355 mm

and

minimum pixel separation at the near point in water

minimum pixel separation that is = 1.22 wavelength / μsin(θ)

minimum pixel separation = 1.22 (550 ×
`10^(-9)`) 25 ×
`10^(-2)` / 1.33 (5) ×
`10^(-3)`

here μ = 1.33

minimum pixel separation = 25.23 ×
`10^(-6)` m

minimum pixel separation = 0.2523 mm

so minimum pixel separation at the near point in water is 0.2523 mm