Question 14 Marks
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
View full question & 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
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-100}-\frac{1}{-25}=\frac{1}{100}=\frac{3}{100}$
$\Rightarrow\text{f}=\frac{100}{3}\text{cm}$
$\text{f}_\text{e}=\frac{100}{3}\text{cm}$
magnification $=\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}=\frac{\big(\frac{200}{3}\big)}{\big( \frac{100}{3}\big)}=2$
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
- With relaxed eye, (normal adjustment)
$\text{f}_\text{e}=\frac{100}{3}\text{cm}$
magnification $=\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}=\frac{\big(\frac{200}{3}\big)}{\big( \frac{100}{3}\big)}=2$