Question

A lady cannot see objects closer than 40 cm from the left eye and closer than 100 cm from the right eye. While on a mountaineering trip, she is lost from her team. She tries to make an astronomical telescope from her reading glasses to look for her teammates. (a) Which glass should she use as the eyepiece? (b) What magnification can she get with relaxed eye?

Answer 1

Given:
For the left glass lens of the lady, we have:
v = – 40 cm and u = – 25 cm
The lens formula is given by
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Putting the values, we get:
$\frac{1}{f}=\frac{1}{\left(-40\right)}-\frac{1}{\left(-25\right)}=\frac{3}{200}$
$\Rightarrow f=\frac{200}{3}=66.6\mathrm{cm}$

For the right glass lens of the lady, we have:
v = – 100 cm, u = – 25 cm
The lens formula is given by
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Putting the values, we get:
$\frac{1}{f}=\frac{1}{\left(-100\right)}-\frac{1}{\left(-25\right)}=\frac{3}{100}$
$\Rightarrow f=\frac{100}{3}=33.3\mathrm{cm}$

(a) An astronomical telescope consists of two lenses: the objective lens having a large focal length and the eyepiece lens having a smaller focal length. So, she should use the right lens of focal length 33.3 cm as the eyepiece lens.

(b) With relaxed eye in normal adjustment,
f_o = 66.6 cm and f_e = 33.3 cm
Magnification $\left(m\right)$ in normal adjustment is given by
m = $\frac{f_0}{f_e}$
$\therefore m=\frac{66.6\mathrm{cm}}{33.3\mathrm{cm}}$= 2

So, with the relaxed eye, she can get the magnification of 2.