A normal human eye can see an object making an angle of ${1.8}^{\circ}$ at the eye. What is the approximately height of object which can be seen by an eye placed at a distance of $1 m$ from the object.

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Solution

Since $θ$ is very small
$tan θ \sim θ$
$θ = \frac{1.8 \times π}{180}$ rad
$\frac{π}{100}$ rad
$θ = \frac{h}{1}, h = \frac{π}{100 m}$
$h = 0.031 m$

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A normal human eye can see an object making an angle of 1.8∘ at the eye. What is the approximately height of object which can be seen by an eye placed at a distance of 1m from the object.

Since θ is very smalltanθ∼θθ=1.8×π180radπ100radθ=h1,h=π100mh=0.031m

A normal human eye can see an object making an angle of ${1.8}^{\circ}$ at the eye. What is the approximately height of object which can be seen by an eye placed at a distance of $1 m$ from the object.

Since $θ$ is very small
$tan θ \sim θ$
$θ = \frac{1.8 \times π}{180}$ rad
$\frac{π}{100}$ rad
$θ = \frac{h}{1}, h = \frac{π}{100 m}$
$h = 0.031 m$