Question

Two identically built microscopes are being used to characterize sub-micron features with two different laser sources. The first microscope uses a laser at 632 nm (red), and the smallest feature it can measure is 500 nm. If the second microscope uses a laser at 543 nm (green), what is the smallest feature it can measure

Answer 1

Answer:

$r_2\approx430\ nm$

Explanation:

Given:

wavelength used by the first microscope, $\lambda_1=632\times 10^{-9}\ m$

resolving power of the first microscope, $r_1=500\times 10^{-9}\ m$

wavelength used by the second microscope, $\lambda_2=543\times 10^{-9}\ m$

For the microscopes the resolving power is given as:

$r=1.22\times \frac{\lambda}{a}$

where:

$\lambda =$ wavelength of the laser source used

$a=$ numerical aperture of the microscope

$r=$ smallest feature that can be distinguished using the optical device

For the first microscope:

$r_1=1.22\times \frac{\lambda_1}{a_1}$

$500\times 10^{-9}=1.22\times\frac{632\times 10^{-9}}{a_1}$

$a_1=1.54208$

Since both the optical device are identically built we will have:

$a_2=a_1=1.54208$

Now for the second optical device:

$r_2=1.22\times \frac{\lambda_2}{a_2}$

$r_2=1.22\times \frac{543\times 10^{-9}}{1.54208}$

$r_2\approx430\ nm$