Solved Problem on Optical Instruments

The objective of a simple telescope has a focal length of 60 cm, and the eyepiece has a focal length of 1.5 cm. The image of an observed star will form 43.5 cm from the eyepiece. Determine the length of the tube of the telescope.

Image const:ruction:

Using the rule, all rays of light incident at the center of the lens continue in a straight line (Figure 1), a ray of light that passes through the optical center of objective O1 does not deviate, being this ray is inclined relative to the principal axis, it will be a secondary axis and, will determine in the focal plane, a secondary focus where the image i1 is formed.

Figure 1

A second ray of light parallel to the secondary axis passes through the lens and is refracted, leaving the secondary focus (Figure 2).

Figure 2

Note: This ray is not necessary to determine the image i1, it illustrates that for objects at infinity, all rays arrive parallel to the instrument and that rays parallel to the secondary axis are refracted by the secondary focus.

The image i1 of the objective is now object o2 for the eyepiece (\( i_{1}\equiv o_{2} \)). Again, a ray that passes through the optical center, this time from the eyepiece O2, is not deviated (Figure 3).

Figure 3

Using the rule, all rays of light parallel to the principal axis pass through the focal point F', and we have a ray that leaves o2 parallel to the principal axis through the focus F'2 (Figure 4).

Figure 4

The two rays found above do not determine an image on the observer's side, and to determine, the image it is necessary to extend these rays to the side of the object o2, and from their crossing, we have the enlarged image i2 (Figure 5).

Figure 5

Problem data:
  • Objective focal length:    f1 = 60 cm;
  • Eyepiece focal length:    f2 = 1.5 cm;
  • Distance from the image to the eyepiece:    p'2 = −43.5 cm.
Problem diagram:

Using the sign convention, on the incident side, we have a positive abscissa for the real object, p > 0, and a negative one for the virtual image, p' < 0, and on the opposite side, we have a negative abscissa of the virtual object, p < 0, and positive for the real image p' > 0.
Image i1 has a positive abscissa, it is behind the objective lens (p'1 > 0, real image), it is also an object for the ocular lens, as it is in front of the lens it has a positive abscissa (p2 > 0, real object), image i2 is in front of the ocular lens, it is a virtual image and has a negative abscissa (p'2 < 0).

Figure 6


The length of the tube will be the sum of the focal length of the objective, f1, with the distance p2, which represents the object to the eyepiece (Figure 6). To find p2, we use the Thin Lens Equation
\[ \begin{gather} \bbox[#99CCFF,10px] {\frac{1}{f}=\frac{1}{p}+\frac{1}{p'}} \end{gather} \]
applying to the second lens (eyepiece)
\[ \begin{gather} \frac{1}{f_{2}}=\frac{1}{p_{2}}+\frac{1}{p'_{2}}\\[5pt] \frac{1}{p_{2}}=\frac{1}{f_{2}}-\frac{1}{p'_{2}}\\[5pt] \frac{1}{p_{2}}=\frac{1}{1.5}-\frac{1}{(-43.5)}\\[5pt] \frac{1}{p_{2}}=\frac{1}{1.5}+\frac{1}{43.5} \end{gather} \]
writing \( 1.5=\frac{15}{10} \) and \( 43.5=\frac{435}{10} \)
\[ \begin{gather} \frac{1}{p_{2}}=\frac{1}{\frac{15}{10}}+\frac{1}{\frac{435}{10}}\\[5pt] \frac{1}{p_{2}}=\frac{10}{15}+\frac{10}{435} \end{gather} \]
multiplying and dividing by 29, the first term on the right side of the equation
\[ \begin{gather} \frac{1}{p_{2}}=\frac{29}{29}\times\frac{10}{15}+\frac{10}{435}\\[5pt] \frac{1}{p_{2}}=\frac{290+10}{435}\\[5pt] \frac{1}{p_{2}}=\frac{300}{435}\\[5pt] p_{2}=\frac{435}{300}\\[5pt] p_{2}=1.45\;\text{cm} \end{gather} \]
the length d of the tube will be
\[ \begin{gather} d=f_{1}+p_{2}\\[5pt] d=60+1.45 \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {d=61.45\;\text{cm}} \end{gather} \]