Question

Assuming the human pupil to have a radius of $0.25\mathrm{cm}$ and a comfortable viewing distance of $25\mathrm{cm}$, the minimum separation between two objects that the human eye can resolve at $500\mathrm{nm}$ wavelength is:

• A. $30\mathrm{\mu m}$
• B. $3\mathrm{\mu m}$
• C. $300\mathrm{\mu m}$
• D. $100\mathrm{\mu m}$

Answer 1

The correct option is A

$30\mathrm{\mu m}$

Step 1: Given information

Radius, $\mathrm{r}=0.25\mathrm{cm}$

Diameter, $$\begin{gathered} \mathrm{d}=2\mathrm{r} \\ =2\times 0.25 \\ =0.5\mathrm{cm} \end{gathered}$$

The wavelength of light, $$\begin{gathered} \mathrm{\lambda }=500\mathrm{nm} \\ =5\times {10}^{-7}\mathrm{m} \end{gathered}$$

Let,

The minimum separation between two objects is $\mathrm{x}$.

Distance, $\mathrm{D}=25\mathrm{cm}$

Step 2: To find.

We have to find the minimum separation between two objects that the human eye can resolve at $500\mathrm{nm}$ wavelength.

Step 3: Calculation

We know the formula,

$\mathrm{sin\theta }=\frac{1.22\mathrm{\lambda }}{\mathrm{d}}$

By substituting the values, we get

$$\begin{gathered} =\frac{1.22\times 5\times {10}^{-7}}{0.5\times {10}^{-2}} \\ =1.22\times {10}^{-4} \end{gathered}$$

Now,

$\mathrm{sin\theta }=\frac{\mathrm{x}}{\mathrm{D}}$

or,

$$\begin{gathered} \mathrm{x}=\mathrm{Dsin\theta } \\ =0.25\times 1.22\times {10}^{-4} \\ =30\times {10}^{-6} \\ =30\mathrm{\mu m} \end{gathered}$$

Therefore, option A is the correct choice.