Question 25.4: Microscope Magnifications Goal Understand the critical facto...
Microscope Magnifications
Goal Understand the critical factors involved in determining the magnifying power of a microscope.
Problem A certain microscope has two interchangeable objectives. One has a focal length of 2.0 \mathrm{~cm}, and the other has a focal length of 0.20 \mathrm{~cm}. Also available are two eyepieces of focal lengths 2.5 \mathrm{~cm} and 5.0 \mathrm{~cm}. If the length of the microscope is 18 \mathrm{~cm}, compute the magnifications for the following combinations: the 2.0-cm objective and 5.0-\mathrm{cm} eyepiece; the 2.0-cm objective and 2.5-cm eyepiece; the 0.20-cm objective and 5.0-cm eyepiece.
Strategy The solution consists of substituting into Equation 25.7
m=M_{1}m_{e}=-{\frac{L}{f_{o}}}{\Biggr(\frac{25\ \mathrm{cm}}{f_{e}}}\Biggr) (25.7)
for three different combinations of lenses.
Apply Equation 25.7 and combine the 2.0-cm objective with the 5.0-cm eyepiece:
m=-\frac{L}{f_{o}}\left(\frac{25 \mathrm{~cm}}{f_{e}}\right)=-\frac{18 \mathrm{~cm}}{2.0 \mathrm{~cm}}\left(\frac{25 \mathrm{~cm}}{5.0 \mathrm{~cm}}\right)=-45
Combine the 2.0-cm objective with the 2.5-cm eyepiece:
m=-\frac{18 \mathrm{~cm}}{2.0 \mathrm{~cm}}\left(\frac{25 \mathrm{~cm}}{2.5 \mathrm{~cm}}\right)=-9.0 \times 10^{1}
Combine the 0.20-\mathrm{cm} objective with the 5.0-\mathrm{cm} eyepiece:
m=-\frac{18 \mathrm{~cm}}{0.20 \mathrm{~cm}}\left(\frac{25 \mathrm{~cm}}{5.0 \mathrm{~cm}}\right)=-450
Remarks Much higher magnifications can be achieved, but the resolution starts to fall, resulting in fuzzy images that don’t convey any details. (See Section 25.6 for further discussion of this point.)