A lady cannot see objects closer than 40cm from the left eye and …

Question

A lady cannot see objects closer than 40cm from the left eye and closer than 100cm from the right eye. While on a mountaineering trip, she is lost from her team. She tries to make an astronomical telscope from her reading glasses to look for her teammates.

Which glass should she use as the eyepiece?
What magnification can she get with relaxed eye?

✓

Answer

The lady can not see objects closer than 40cm from the left eye and 100cm from the right eye.For the left glass lens,
v = -40cm,
u = -25cm
$\therefore \frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-40}-\frac{1}{-25}=\frac{1}{25}-\frac{1}{40}=\frac{3}{200}$
$\Rightarrow \text{f}=\frac{200}{3}\text{cm}$
For the right glass lens,
v = -100cm,
u = -25cm
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-100}-\frac{1}{-25}=\frac{1}{100}=\frac{3}{100}$
$\Rightarrow\text{f}=\frac{100}{3}\text{cm}$

For an astronomical telescope, the eye piece lens should have smaller focal length. So, she should

use the right lens $\big(\text{f}=\frac{100}{3}\text{cm}\big)$ as the eye piece lens.

With relaxed eye, (normal adjustment)

$\text{f}_0=\frac{200}{3}\text{cm}$
$\text{f}_\text{e}=\frac{100}{3}\text{cm}$
magnification $=\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}=\frac{\big(\frac{200}{3}\big)}{\big( \frac{100}{3}\big)}=2$

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