Respuesta :
Answer:
[tex]r_2\approx430\ nm[/tex]
Explanation:
Given:
wavelength used by the first microscope, [tex]\lambda_1=632\times 10^{-9}\ m[/tex]
resolving power of the first microscope, [tex]r_1=500\times 10^{-9}\ m[/tex]
wavelength used by the second microscope, [tex]\lambda_2=543\times 10^{-9}\ m[/tex]
For the microscopes the resolving power is given as:
[tex]r=1.22\times \frac{\lambda}{a}[/tex]
where:
[tex]\lambda =[/tex] wavelength of the laser source used
[tex]a=[/tex] numerical aperture of the microscope
[tex]r=[/tex] smallest feature that can be distinguished using the optical device
For the first microscope:
[tex]r_1=1.22\times \frac{\lambda_1}{a_1}[/tex]
[tex]500\times 10^{-9}=1.22\times\frac{632\times 10^{-9}}{a_1}[/tex]
[tex]a_1=1.54208[/tex]
Since both the optical device are identically built we will have:
[tex]a_2=a_1=1.54208[/tex]
Now for the second optical device:
[tex]r_2=1.22\times \frac{\lambda_2}{a_2}[/tex]
[tex]r_2=1.22\times \frac{543\times 10^{-9}}{1.54208}[/tex]
[tex]r_2\approx430\ nm[/tex]
