Chapter 25, Problem 37P

To determine

The maximum achievable angular magnification for a telescope designed to be hand held.

Answer to Problem 37P

Solution:

The maximum angular magnification possible for a hand held astronomical telescope is $-9\times$

Explanation of Solution

If a lens is used as a magnifying glass, the object is placed between the focus and the optic centre and the image is formed at the near point N of the eye, which is 25 cm for a normal eye. For a relaxed eye, the object distance may be taken equal to that of the focal length of the lens $f_e$.

The magnification $m_e$ of the eyepiece is given by the expression,

$m_e=\frac{N}{f_e}$

The length L of a telescope is the sum of the focal lengths of the objective $f_o$ and the eye piece $f_e$.

$L=f_o+f_e$

The focal length of the objective can be calculated using the above expression and the magnificationM of the telescope is given by the ratio of the focal lengths of the objective and the eye piece.

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

Given:

The maximum length of the telescope $L=50\text{ cm}$

Magnification of the eyepiece $M=+5\times$

Near point of the eye $N=25\text{ cm}$

Formula:

$m_e=\frac{N}{f_e}$

$L=f_o+f_e$

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

Calculation:

Using the expression $m_e=\frac{N}{f_e}$ and the value of the near point as 25 cm and the magnification of the eye lens as +5, calculate the focal length of the eye lens.

$\begin{array}{c}{f}_e=\frac{N}{m_e}=\frac{25\text{ cm}}{5}\\ =5\text{ cm}\end{array}$

Using the expression $L=f_o+f_e$ and the calculated value of $f_e$ and the given value of L, the focal length of the objective can be calculated.

$\begin{array}{c}{f}_o=L-f_e\\ =50\text{ cm}-5\text{ cm}\\ =45\text{ cm}\end{array}$

The magnification of the telescope can be calculated using the calculated values of $f_e$ and $f_o$ in the expression,

$\begin{array}{c}M=-\frac{f_o}{f_e}\\ =-\frac{45\text{ cm}}{5\text{ cm}}\\ =-9\times \end{array}$