A compound microscope has a barrel whose length is 17.4 cm and an eyepiece whose focal length is 1.7 cm. The viewer has a near point located 25 cm from his eyes. What focal length must the objective have so the angular magnification of the microscope is -330?
To find the focal length of the objective of the compound microscope, we can use the formula for angular magnification:
\(M = - \frac{D_{o}}{D_{e}}\)
Where:
M = Angular magnification
\(D_{o}\) = Distance between the objective lens and the image
\(D_{e}\) = Distance between the viewer's eyes and the image
In this case, we are given:
\(M = -330\)
\(D_{e} = 25\) cm
\(D_{o} = L - D_{e}\)
Where L is the total length of the microscope, which is the sum of the barrel length and the focal length of the eyepiece.
To find the focal length of the objective, we need to solve for \(D_{o}\) and substitute the given values into the formula.
Step 1: Find L
\(L = \text{barrel length} + \text{focal length of eyepiece}\)
\(L = 17.4 \, \text{cm} + 1.7 \, \text{cm}\)
\(L = 19.1 \, \text{cm}\)
Step 2: Find \(D_{o}\)
\(D_{o} = L - D_{e}\)
\(D_{o} = 19.1 \, \text{cm} - 25 \, \text{cm}\)
\(D_{o} = -5.9 \, \text{cm}\)
Step 3: Substitute the values into the formula for angular magnification
\(M = - \frac{D_{o}}{D_{e}}\)
\(-330 = - \frac{-5.9 \, \text{cm}}{25 \, \text{cm}}\) (Note: We use the negative value for \(D_{o}\) as stated in the problem statement.)
\(-330 = 0.236\, \text{(approximately)}\)
Rearranging the formula to solve for \(D_{o}\) gives:
\(D_{o} = -330 \times D_{e}\)
\(D_{o} = -330 \times 25 \, \text{cm}\)
\(D_{o} = -8250 \, \text{cm}\)
Now, we can find the focal length of the objective using \(D_{o}\):
\(D_{o} = f_{o} - L\)
Rearranging this formula to solve for \(f_{o}\) gives:
\(f_{o} = D_{o} + L\)
\(f_{o} = -8250 \, \text{cm} + 19.1 \, \text{cm}\)
\(f_{o} = -8230.9 \, \text{cm}\)
So, the focal length of the objective must be approximately -8230.9 cm for the angular magnification of the microscope to be -330.