Question

A microscope with an objective of focal length 1.8 mm is used to inspect the tiny features of a computer chip. It is desired to resolve two objects only 400 nm apart. What diameter objective is needed if the microscope is used in air with light of wavelength 550 nm?

 

Answer 1

You have to use the formula:

\(\mathrm{D}=(2)(\mathrm{f})\left(\right.\) tan (inverse of \(\sin \left((0.61 \mathrm{lamba}) /\left(\mathrm{d}^{*} \mathrm{n}\right)\right)\)

So, for this problem, you would work it out like this:

\(\mathrm{D}=(2)(1.8 \mathrm{~mm})\) tan(inverse of \(\sin \left(\left(0.61\left(550 \times 10^{-6} \mathrm{~mm}\right) /\left(400^{*} 10^{-6}\right)(1)\right)\right.\)

\(\mathrm{D}=(2)(1.8 \mathrm{~mm})\) tan(inverse of \(\sin \left(\left(8.3875^{*} 10^{-1}\right)\right)\)

\(\mathrm{D}=(2)(1.8 \mathrm{~mm}) \tan \left(5.7008357^{*} 10^{1} \right)\)

\(D=(2)(1.8 m m)\left(1.540356795 * 10^{0} \right)\)

\(D=5.54\)