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


  1. Answer:

    r_2\approx430\ nm



    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}


    \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}


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


    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

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