Question 25.6: Resolution of a Microscope Goal Study limitations on the res...
Resolution of a Microscope
Goal Study limitations on the resolution of a microscope.
Problem Sodium light of wavelength 589 \mathrm{~nm} is used to view an object under a microscope. The aperture of the objective has a diameter of 0.90 \mathrm{~cm}. (a) Find the limiting angle of resolution for this microscope. (b) Using visible light of any wavelength you desire, find the maximum limit of resolution for this microscope. (c) Water of index of refraction 1.33 now fills the space between the object and the objective. What effect would this have on the resolving power of the microscope, using 589 \mathrm{~nm} light?
Strategy Parts (a) and (b) require substitution into Equation 25.10.
\theta_{\mathrm{min}}=1.22{\frac{\lambda}{D}} (25.10)
Because the wavelength appears in the numerator, violet light, with the shortest visible wavelength, gives the maximum resolution. In part (c), the only difference is that the wavelength changes to \lambda / n, where n is the index of refraction of water.
(a) Find the limiting angle of resolution for this microscope.
Substitute into Equation 25.10 to obtain the limiting angle of resolution:
\begin{aligned} \theta_{\text {min }} & =1.22 \frac{\lambda}{D}=1.22\left(\frac{589 \times 10^{-9} \mathrm{~m}}{0.90 \times 10^{-2} \mathrm{~m}}\right) \\ & =8.0 \times 10^{-5} \mathrm{rad} \end{aligned}
(b) Calculate the microscope’s maximum limit of resolution.
To obtain the maximum resolution, substitute the shortest visible wavelength available — violet light, of wavelength 4.0 \times 10^{2} \mathrm{~nm} :
\begin{aligned} \theta_{\text {min }} & =1.22 \frac{\lambda}{D}=1.22\left(\frac{4.0 \times 10^{-7} \mathrm{~m}}{0.90 \times 10^{-2} \mathrm{~m}}\right) \\ & =5.4 \times 10^{-5} \mathrm{rad} \end{aligned}
(c) What effect does water between the object and the objective lens have on the resolution, with 589-nm light?
Calculate the wavelength of the sodium light in the water:
\lambda_{w}=\frac{\lambda_{a}}{n}=\frac{589 \mathrm{~nm}}{1.33}=443 \mathrm{~nm}
Substitute this wavelength into Equation 25.10 to get the resolution:
\theta_{\text {min }}=1.22\left(\frac{443 \times 10^{-9} \mathrm{~m}}{0.90 \times 10^{-2} \mathrm{~m}}\right)=6.0 \times 10^{-5} \mathrm{rad}
Remarks In each case, any two points on the object subtending an angle of less than the limiting angle \theta_{\min } at the objective cannot be distinguished in the image. Consequently, it may be possible to see a cell, but then be unable to clearly see smaller structures within the cell. Obtaining an increase in resolution is the motivation behind placing a drop of oil on the slide for certain objective lenses.