Chapter 25, Problem 36P

To determine

The focal lengths of the objective and eye lenses of the given astronomical telescope.

Answer to Problem 36P

Solution:

The focal length of the objective lens is found to be 1.09 m and that of the eye lens is found to be 0.91cm.

Explanation of Solution

Angular magnification M of the telescope is the ratio of the focal lengths of the objective and the eye piece.

$M=-\frac{f_o}{f_e}$

Here, the focal length of the objective lens is $f_o$ and the focal length of the eye piece is $f_e$.

The length l of the telescope is the sum of the focal lengths of the objective and the eye piece.

$l=f_o+f_e$

Using these expressions, the focal lengths of the objective and the eyepiece can be calculated.

Given:

The length of the telescope $l=1.10\text{ m}$

The magnification of the telescope $M=120\times$

Formula:

$\begin{array}{l}M=-\frac{f_o}{f_e}\\ l=f_o+f_e\end{array}$

Calculation:

Use the given quantities in the above expressions to get equations for $f_o$ and $f_e$. The image formed by an astronomical telescope is inverted, hence affix a negative sign to the magnification.

$\begin{array}{c}{f}_o=-M\times f_e\\ =-\left(-120\right)f_e\\ =120f_e\end{array}$

Using the value for length,

$f_o+f_e=1.10\text{ m}$

Solve the equations $f_o+f_e=1.10\text{ m;}{f}_o=120f_e$ to obtain the values of $f_o$ and $f_e$.

$\begin{array}{c}120f_e+f_e=1.10\text{ m}\\ f_e=\frac{1.10\text{ m}}{120}\\ =9.09\times {10}^{-3}\text{m}\\ =0.91\text{ cm}\end{array}$

Substitute the value of $f_e$ in the equation $f_o+f_e=1.10\text{ m}$ and calculate the value of $f_o$.

$\begin{array}{c}{f}_o+0.909\text{ cm}=1.10\text{ m}\\ f_o=110\text{ cm}-0.909\text{ cm}\\ =109.091\text{ cm}\\ =1.09\text{ m}\end{array}$