A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Exercise
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Focal length of the objective lens, $f_1 = 2.0$ cm
Focal length of the eyepiece, $f_2 = 6.25$ cm
Distance between the objective lens and the eyepiece, d = 15 cm
Object distance for the eyepiece $= u_2$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}_2}$
$\frac{1}{\text{u}_2}=\frac{1}{\text{v}_2}-\frac{1}{\text{f}_2}$
$=\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$
$\therefore \ \text{u}_2=-5 \ \text{cm}$
Image distance for the objective lens, $\text{v}_1=\text{d}+\text{u}_2=15-5=10 \ \text{cm}$
Object distance for the objective lens $= u_2$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\frac{1}{\text{u}_1}=\frac{1}{\text{v}_1}-\frac{1}{\text{f}_1}$
$=\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}$
$\therefore \ \text{u}_1=-2.5\ \text{cm}$
Magnitude of the object distance, $|u_1| = 2.5$ cm
The magnifying power of a compound microscope is given by the relation:
$\text{m}=\frac{\text{v}_1}{|\text{u}_1|}\Big(1+\frac{\text{d}'}{\text{f}_2}\Big)$
$=\frac{10}{2.5}\Big(1+\frac{25}{6.25}\Big)=4(1+4)=20$
Hence, the magnifying power of the microscope is 20.
Object distance for the eyepiece $= u_2$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}_2}$
$\frac{1}{\infty}-\frac{1}{\text{u}_2}=\frac{1}{6.25}$
$\therefore \ \text{u}_2=-6.25\ \text{cm}$
Image distance for the objective lens, $\text{v}_1=\text{d}+\text{u}_2=15-6.25=8.75 \ \text{cm}$
Object distance for the objective lens $= u_1$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\frac{1}{\text{u}_1}=\frac{1}{\text{v}_1}-\frac{1}{\text{f}_1}$
$=\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}$
$\therefore \ \text{u}_1=-\frac{17.5}{6.75}=-2.59\ \text{cm}$
Magnitude of the object distance, $u_1 = 2.59$ cm
The magnifying power of a compound microscope is given by the relation:
$\text{m}=\frac{\text{v}_1}{|\text{u}_1|}\Bigg(\frac{\text{d}'}{|\text{u}_2|}\Bigg)$
$=\frac{8.75}{2.59}\times\frac{25}{6.25}=13.51$
Hence, the magnifying power of the microscope is 13.51.
Focal length of the eyepiece, $f_2 = 6.25$ cm
Distance between the objective lens and the eyepiece, d = 15 cm
- Least distance of distinct vision, d' = 25 cm
Object distance for the eyepiece $= u_2$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}_2}$
$\frac{1}{\text{u}_2}=\frac{1}{\text{v}_2}-\frac{1}{\text{f}_2}$
$=\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$
$\therefore \ \text{u}_2=-5 \ \text{cm}$
Image distance for the objective lens, $\text{v}_1=\text{d}+\text{u}_2=15-5=10 \ \text{cm}$
Object distance for the objective lens $= u_2$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\frac{1}{\text{u}_1}=\frac{1}{\text{v}_1}-\frac{1}{\text{f}_1}$
$=\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}$
$\therefore \ \text{u}_1=-2.5\ \text{cm}$
Magnitude of the object distance, $|u_1| = 2.5$ cm
The magnifying power of a compound microscope is given by the relation:
$\text{m}=\frac{\text{v}_1}{|\text{u}_1|}\Big(1+\frac{\text{d}'}{\text{f}_2}\Big)$
$=\frac{10}{2.5}\Big(1+\frac{25}{6.25}\Big)=4(1+4)=20$
Hence, the magnifying power of the microscope is 20.
- The final image is formed at infinity.
Object distance for the eyepiece $= u_2$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}_2}$
$\frac{1}{\infty}-\frac{1}{\text{u}_2}=\frac{1}{6.25}$
$\therefore \ \text{u}_2=-6.25\ \text{cm}$
Image distance for the objective lens, $\text{v}_1=\text{d}+\text{u}_2=15-6.25=8.75 \ \text{cm}$
Object distance for the objective lens $= u_1$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\frac{1}{\text{u}_1}=\frac{1}{\text{v}_1}-\frac{1}{\text{f}_1}$
$=\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}$
$\therefore \ \text{u}_1=-\frac{17.5}{6.75}=-2.59\ \text{cm}$
Magnitude of the object distance, $u_1 = 2.59$ cm
The magnifying power of a compound microscope is given by the relation:
$\text{m}=\frac{\text{v}_1}{|\text{u}_1|}\Bigg(\frac{\text{d}'}{|\text{u}_2|}\Bigg)$
$=\frac{8.75}{2.59}\times\frac{25}{6.25}=13.51$
Hence, the magnifying power of the microscope is 13.51.
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