Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm . Approxim

Question

Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm . Approximately, what is the maximum distance at which these dots can be resolved by the eye? [Take wavelength of light = 500 nm ]

Accepted Answer

According to the Rayleigh criterion for resolution of the eye:Angular resolution

θ = \frac{1.22 λ}{d}

Also,

θ = \frac{x}{D}

, where

x

is the separation between dots and

D

is the distance from the eye.

\frac{x}{D} = \frac{1.22 λ}{d} ⟹ D = \frac{x \cdot d}{1.22 λ}

Given

x = 1 mm = 1 0^{- 3} m, d = 3 mm = 3 \times 1 0^{- 3} m, λ = 500 nm = 5 \times 1 0^{- 7} m

:

D = \frac{1 0 ^{- 3} \cdot 3 \times 1 0 ^{- 3}}{1.22 \cdot 5 \times 1 0 ^{- 7}}

D = \frac{3 \times 1 0 ^{- 6}}{6.1 \times 1 0 ^{- 7}} = \frac{30}{6.1} \approx 4.92 m

Approximately

D = 5 m

.

Two point white dots are $1 mm$ apart on a black paper. They are viewed by eye of pupil diameter $3 mm$ . Approximately, what is the maximum distance at which these dots can be resolved by the eye? [Take wavelength of light $= 500 nm$ ]

Solution & Step-by-step Explanation

Practice this question

Share This Question

Related Questions

Ready for a Full Test?

Discussion

Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [Take wavelength of light =500 nm ]