Q61E - University Physics with Modern Physics

Question

A telescope is constructed from two lenses with focal lengths of 95.0 cm and 15.0 cm, the 95.0 cm lens is being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building 60.0 m tall, 3.00 km away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

Step-by-Step Solution

Answer

1. The angular magnification for the telescope is -6.33.
2. The height of the image formed by the objective of a building is 1.9 cm.
3. The angular size of the final image as viewed by an eye very close to the eyepiece is -0.127 rad.

Step 1: Definition of angular magnification of the telescope and lateral magnification and formula of angular size of the object and angular size of the image

The angular magnification of the telescope is the ratio between focal length of the object and focal length of the eyepiece.

                              $M=-\frac{f_1}{f_2}$

The lateral magnification is the ratio between the distance of the object and the distance of the image.

                             $m=-\frac{s^{\text{'}}}{s}$

Formula for angular size of the object: $\theta =\frac{y^{\text{'}}}{f_2}$

Where y' =  height of the image

Formula for angular size of the image:  ${\theta }^{\text{'}}=\theta M$

Step 2: Calculate the angular magnification

The telescope has the same principle as the microscope but for a very long distance object. 

Plugging the values into the angular magnification formula:

 $$\begin{gathered} M=-\frac{f_1}{f_2} \\ =-\frac{95.0cm}{15.0cm} \\ =-6.33 \end{gathered}$$

Step 3: Calculate the lateral magnification

Use the formula, 

                                    $$\begin{gathered} m=-\frac{s^{\text{'}}}{s} \\ =-\frac{95m}{3*{10}^{5}cm} \\ =-3.17*{10}^{-4} \end{gathered}$$

Also, the lateral magnification is given by: 

                                       $$\begin{gathered} m=-\frac{y^{\text{'}}}{y} \\ \Rightarrow \left|y^{\text{'}}\right|=\left|my\right| \\ =\left|\left(-3.17*{10}^{-4}\right)\left(60 ml\right)\right| \\ =0.019 \\ =1.9cm \end{gathered}$$

 

Step 4: Calculate angular size of the final image

First, calculate the angular size of the object

                                     $$\begin{gathered} \theta =\frac{y^{\text{'}}}{f_2} \\ =\frac{1.9cm}{95cm} \\ =0.02rad \end{gathered}$$

Plug the values to find angular size of the final image

                                         $$\begin{gathered} {\mathrm{\theta }}^{\text{'}}=\mathrm{\theta M} \\ =\left(0.02\mathrm{rad}\right)\left(-6.33\right) \\ =-0.127\mathrm{rad} \end{gathered}$$

Thus, The angular magnification for the telescope is -6.33, the height of the image formed by the objective of a building is 1.9 cm and the angular size of the final image as viewed by an eye very close to the eyepiece is -0.127 rad.