Question

If you can read the bottom row of your doctor's eye chart, your eye has a resolving power of one arcminute, equal to 1.67E-2 degrees. If this resolving power is diffraction-limited, to what effective diameter of your eye's optical system does this corresponding? Use Rayleigh's criterion and assume that the wavelenght of the light is 555nm.

Answer 1

Concepts and reason

The concept used to solve this question is Rayleigh criterion.

First, convert the value of $\theta$from degree to radians. Use the value of wavelength and $\theta$in the expression of effective diameter.

Fundamentals

The Rayleigh criteria states that the condition of minimum resolution.

The expression of the Rayleigh criteria is given as follows:

$1.22\lambda = D\theta$

Here, $\lambda$ is the wavelength of light, D is the effective diameter of eye’s optical system, and $\theta$ is the small angle in radians.

Convert the value of $\theta$from degree to radians.

$\begin{array}{c}\\\theta = 0.0167^\circ \left( {\frac{{3.14{\rm{ radians}}}}{{180^\circ }}} \right)\\\\ = 0.000291{\rm{ radians}}\\\end{array}$

The expression for the Rayleigh criteria is given by,

$1.22\lambda = D\theta$

Rearrange the above expression for D as follows:

$D = \frac{{1.22\lambda }}{\theta }$

Convert the value of $\lambda$from nm to m.

$\begin{array}{c}\\\lambda = 555{\rm{ nm}}\left( {\frac{{{{10}^{ - 9}}{\rm{ m}}}}{{1.0{\rm{ nm}}}}} \right)\\\\ = 555 \times {10^{ - 9}}{\rm{ m}}\\\end{array}$

Substitute $555 \times {10^{ - 9}}{\rm{ m}}$for $\lambda$and 0.000291 radians for $\theta$in the equation$D = \frac{{1.22\lambda }}{\theta }$.

$\begin{array}{c}\\D = \frac{{1.22\left( {555 \times {{10}^{ - 9}}{\rm{ m}}} \right)}}{{0.000291{\rm{ radians}}}}\\\\ = 2.33 \times {10^{ - 3}}{\rm{ m}}\\\end{array}$

Ans:

The value of the effective diameter is $2.33 \times {10^{ - 3}}{\rm{ m}}$.