Question

Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two object that human eye can resolve at 500 nm wavelength is :
A. $1\mu m$
B. $30\mu m$
C. $100\mu m$
D. $300\mu m$

Answer 1

Hint: The angular separation is the ratio of minimum separation and distance from the eye and angular separation is also equal to 1.22 times the wavelength divided by diameter of the pupil.

Complete step by step answer:A radius (r) of the human pupil is $0.25cm\left( {0.25 \times {{10}^{ - 2}}m} \right)$. The comfortable viewing distance from the human eye (D) is 25 cm. The wavelength (λ) is given as 500 nm.
The resolving power of a lens is defined as the ability of the lens to differentiate between two lines or points (or two objects). The resolving power is inversely proportional to the distance between two objects i.e., greater is the resolving power, smaller is distance between them.
The resolving power (R) is reciprocal of the angular separation $\left( \theta \right)$ between two distant objects i.e., $R = \dfrac{1}{\theta }$.
Now, $\theta = \dfrac{{1.22\lambda }}{d}$where d is the diameter of the lens i.e., pupil
$\theta = \dfrac{{1.22 \times 500 \times {{10}^{ - 9}}}}{{2r}}\left[ {1nm = {{10}^{ - 9}}m,d = 2r} \right]$
$\Rightarrow \theta = \dfrac{{610 \times {{10}^{ - 9}}}}{{2 \times 0.25 \times {{10}^{ - 2}}}}$
$\Rightarrow \theta = 1220 \times {10^{ - 7}}$
$\Rightarrow \theta = 1.22 \times {10^{ - 4}}$
Now, let minimum separation between two objects be x.
Thus, $\theta = \dfrac{x}{D}$
$1.22 \times {10^{ - 4}} = \dfrac{x}{{25 \times {{10}^{ - 2}}}}$
$\Rightarrow x = 1.22 \times {10^{ - 4}} \times 25 \times {10^{ - 2}}$
$\Rightarrow x = 30.5 \times {10^{ - 6}}$
$\Rightarrow x = 30\mu m$
Therefore, option B is correct.

Note:The value 1.22 applies to circular apertures like the pupil of our eye or the apertures in telescopes, microscopes or cameras. The diameter of the reflecting surface of a spherical mirror is called aperture.