Question

The eyepiece of a refracting telescope has a focal length of 9.00 cm. The distance between objective and eyepiece is 1.8 m, and the final image is at infinity. What is the angular magnification of the telescope?

Answer 1

Concepts and reason

The required concepts to solve this question are angular magnifiction and distnce between object and eyepiece.

Firstly, write the equation for distance. Rearrange the equation for focal length of object and calculate the angular magnification of the telescope.

Fundamentals

The expression for angular magnification is,

${M_\alpha } = - \frac{{{f_o}}}{{{f_e}}}$

Here, ${f_o}$is the focal length of the eyepiece and ${f_e}$ is the focal length of the object.

The expression for distance between object and eyepiece is,

$d = {f_o} + {f_e}$

Distance between object and eyepiece is $1.8{\rm{ m}}$.

The expression for total distance is,

$d = {f_o} + {f_e}$

Substitute $1.8{\rm{ m}}$ for $d$and $9{\rm{ cm}}$ for ${f_e}$ in above equation.

$\begin{array}{c}\\1.8{\rm{ m}} = {f_o} + 9{\rm{ cm}}\\\\{f_o}{\rm{ = }}1.8{\rm{ m}}\left( {\frac{{{{10}^2}{\rm{ cm}}}}{{{\rm{ 1 m}}}}} \right) - 9{\rm{ cm}}\\\\{\rm{ = 171}}\,{\rm{cm}}\\\end{array}$

The expression for angular magnification is,

${M_\alpha } = - \frac{{{f_o}}}{{{f_e}}}$

Substitute ${\rm{171}}\,{\rm{cm}}$ for ${f_o}$and $9{\rm{ cm}}$ for ${f_e}$ in above equation.

$\begin{array}{c}\\{M_\alpha } = - \frac{{{\rm{171}}\,{\rm{cm}}}}{{9{\rm{ cm}}}}\\\\ = - 19\\\end{array}$

Ans:

The angular magnification of the telescope is $- 19$.