Chapter 38, Problem 80PQ

CASE STUDY A group of students is given two converging lenses. Lens A has a focal length of 12.5 cm. and lens B has a focal length of 50.0 cm. The diameter of each lens is 6.50 cm. The students are asked to construct a telescope from these lenses if possible, and they have this discussion:

Avi: To make a telescope, we pick lens B to be the objective and lens A to be the eyepiece. Lens B has the greater focal length, so it has to be the objective.

Cameron: Both lenses have the same diameter—6.50 cm. It doesn’t matter which is the objective.

Shannon: It does matter because the magnification depends on their relative focal lengths. We still want to get the best magnification.

1. a. What do you think?
2. b. If a telescope can be constructed from these two lenses, describe its design. What are its LGP and angular magnification? Compare the LGP to the value for your fully open pupil.

To determine

The most appropriate argument.

Answer to Problem 80PQ

The statement of the Avi and Shannon is perfectly correct and the statement of Cameron is partially correct.

Explanation of Solution

The telescope is an instrument which is used to magnify the image paced at a larger distance. It has two converging lens, in which one is used as an objective lens and other is used as an eyepiece. The light rays from the distinct object converge at the focal length of the objective lens and then the eyepiece magnifies that image.

In general, the telescope has a large focal length and diameter of the objective lens, and small focal length and small diameter of the eyepiece.

Write the expression to obtain the magnification of a telescope.

    $m_{\text{tel}}=\frac{f_{\text{o}}}{f_{\text{e}}}$

Here, $m_{\text{tel}}$ is the telescope total angular magnification, $f_{\text{o}}$ is the focal length of the objective lens and $f_{\text{e}}$ is the focal length of the eyepiece.

Based on the above expression the magnification of the telescope is inversely proportional to the focal length of the eyepiece.

Thus, to obtain the higher magnification of the telescope, the focal length of the eyepiece should be minimum.

Write the expression to obtain the light gather power of the telescope.

    $LG{P}_{\text{tele}}=d_{\text{obj}}^2$

Here, $LG{P}_{\text{tele}}$ is the light gather power of the telescope and $d_{\text{obj}}$ is the diameter of the objective lens.

The light gathering power of a telescope has higher importance as compared to the magnification of the telescope as light gathering power contributes to the image resolution.

The light gathering power is directly proportional to the square of the diameter of the objective lens.

Therefore, the statement of the Avi and Shannon is perfectly correct and the statement of Cameron is partially correct.

To determine

The design of the telescope, light gathering power of the telescope and the angular magnification and compare the light gathering power of a telescope to the value for the fully open pupil.

Answer to Problem 80PQ

The lens $B$ is used as a objective lens and lens $A$ is used as a eye piece of a telescope, the angular magnification of a telescope is $4.0$, the light gathering power of telescope is $4225{\text{mm}}^2$ and the light gathering power of a telescope is $66$ times higher than the light gathering power of a fully open pupil.

Explanation of Solution

As the objective lens should have higher focal length and the eye piece should have least focal length value. Therefore, the lens $B$ should be objective lens and lens $A$ should be eyepiece.

Write the expression to obtain the magnification of a telescope.

    $m_{\text{tel}}=\frac{f_{\text{o}}}{f_{\text{e}}}$                                                                                                                 (I)

Here, $m_{\text{tel}}$ is the telescope total angular magnification, $f_{\text{o}}$ is the focal length of the objective lens and $f_{\text{e}}$ is the focal length of the eyepiece.

Write the expression to obtain the light gather power of the telescope.

    $LG{P}_{\text{tele}}=d_{\text{obj}}^2$                                                                                                         (II)

Here, $LG{P}_{\text{tele}}$ is the light gather power of the telescope and $d_{\text{obj}}$ is the diameter of the objective lens.

Write the expression to obtain the light gather power of fully open pupil.

    $LG{P}_{\text{eye}}=d_{\text{eye}}^2$                                                                                                        (III)

Here, $LG{P}_{\text{eye}}$ is the light gather power of fully open pupil and $d_{\text{eye}}$ is the diameter of the eye lens.

Conclusion:

Substitute $50.0\text{cm}$ for $f_{\text{o}}$ and $12.5\text{cm}$ for $f_{\text{e}}$ in equation (I) to calculate $m_{\text{tel}}$.

    $\begin{array}{c}{m}_{\text{tel}}=\frac{50.0\text{cm}}{12.5\text{cm}}\\ =4.0\end{array}$

Substitute $6.50\text{cm}$ for $d_{\text{obj}}$ in equation (II) to calculate $LG{P}_{\text{tele}}$.

    $\begin{array}{c}LG{P}_{\text{tele}}={\left(6.50\text{cm}\right)}^2\\ ={\left(6.50\text{cm}\times \frac{10\text{mm}}{1\text{cm}}\right)}^2\\ ={\left(65\text{mm}\right)}^2\\ =4225{\text{mm}}^2\end{array}$

Substitute $8.0\text{mm}$ for $d_{\text{eye}}$ in equation (III) to calculate $LG{P}_{\text{eye}}$.

    $\begin{array}{c}LG{P}_{\text{eye}}={\left(8.0\text{mm}\right)}^2\\ =64{\text{mm}}^2\end{array}$

The ratio of power of a telescope to the power of a fully open pupil.

    $\begin{array}{c}\frac{LG{P}_{\text{tele}}}{LG{P}_{\text{eye}}}=\frac{4225{\text{mm}}^2}{64{\text{mm}}^2}\\ =66.0\end{array}$

Therefore, the lens $B$ is used as a objective lens and lens $A$ is used as a eye piece of a telescope, the angular magnification of a telescope is $4.0$, the light gathering power of telescope is $4225{\text{mm}}^2$ and the light gathering power of a telescope is $66$ times higher than the light gathering power of a fully open pupil.