Question

The focal length of the eyepiece of a certain microscope is 18.0 mm. The focal length of the objective is 8.00 mm. The distance between objective and eyepiece is 19.7 cm. The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?

Short answer

1. The distance from the objective to the object being viewed is 8.374 mm.
2. The magnitude of linear magnification is -21.375.
3. The overall angular magnification of the microscope is -296.875.a

Step-by-step solution

Object-Image Relationship, sign rules for the variables, and Lateral magnification for a thin lens

$\frac{1}{s}+\frac{1}{s^{\text{'}}}=\frac{1}{f}$

Where s = object distance from the lens

s' = image distance from the lens

f = Focal length of the lens

Sign rules for the variables:

1. Sign rule for the object distance: when the object is on the same side of the refracting surface as the incoming light, object distance s is positive; otherwise, it is negative.
2. Sign rule for the image distance: when the image is on the same side of the refracting surface as the outgoing light, image distance s' is positive; otherwise, it is negative.

Lateral magnification for a thin lens:

role="math" localid="1663918992136" $m=\frac{-s^{\text{'}}}{s}=\frac{y^{\text{'}}}{y}$

Where the magnification, object distance, image distance, height of the image, and height of the object

$\mathit{m}\mathbf{\to }$is when the image is erect and (-) when the image is inverted.

Calculate the distance from the objective to the object being viewed

The eyepiece: The image is formed at infinity${s^{\text{'}}}_{eyepiece}=\infty$

The object of the eyepiece is on the focal of the eye piece${s^{\text{'}}}_{eyepiece}=18mm=1.8cm$

The objective: The image of the objective is the object of the eyepiece

$$\begin{gathered} {s^{\text{'}}}_{objective}=19.7cm-1.8cm \\ =17.9cm \\ \Rightarrow f_{objective}=8mm \\ =0.8cm \end{gathered}$$

Use the formula,

$$\begin{gathered} \frac{1}{s}+\frac{1}{s^{\text{'}}}=\frac{1}{f} \\ \Rightarrow \frac{1}{s}=\frac{1}{f}-\frac{1}{s^{\text{'}}} \\ =\frac{1}{0.8}-\frac{1}{17.9} \\ =\frac{855}{716cm} \\ \Rightarrow s_{objective}=\frac{716cm}{855} \\ =0.8374cm \\ =8.374mm \end{gathered}$$

Calculate the magnitude of linear magnification

$$\begin{gathered} m=\frac{-s^{\text{'}}}{s} \\ =\frac{-17.94cm}{0.8374} \\ =-21.375 \end{gathered}$$

Calculate the overall angular magnification of the microscope

$$\begin{gathered} M=M_{angular\_for\_eyepiece}*m \\ =\frac{25cm}{f_1}*-21.375 \\ =\frac{25cm}{1.8cm}*-21.375 \\ =-296.875 \end{gathered}$$

Thus, the distance from the objective to the object being viewed is 8.374 mm, the magnitude of linear magnification is -21.375, and the overall angular magnification of the microscope is -296.875.