Question

A lady cannot see objects closer than $40\text{cm}$ from the left eye and closer than $100\text{cm}$ from the right eye. On a mountaineering trip, she is lost from her team. She tries to make an astronomical telescope from her reading glass. What is the magnification of this telescope for a relaxed eye?

• A. $5$
• B. $4$
• C. $3.5$
• D. $2$

Answer 1

The correct option is D $2$

Given:
$D_{right}=100\text{cm}$ and $D_{left}=40\text{cm}$.

Since, the near point for the eyes of the lady are shifted to a distance more than $25\text{cm}$, we can say that the lady is suffering from far sightedness.

To treat farsightedness convex lenses are used. These lenses form the image of the object lying at the normal near point$\left(25\text{cm}\right)$, at the near point of the defected eye.

Hence, the focal length of the lens used for the right eye:

$\frac{1}{f_{right}}=\frac{1}{25}-\frac{1}{100}$

$\Rightarrow \frac{1}{f_{right}}=\frac{3}{100}$

$\therefore f_{right}=\frac{100}{3}\text{cm}$

Focal length of the lens used for the left eye:

$\frac{1}{f_{left}}=\frac{1}{25}-\frac{1}{40}$

$\Rightarrow f_{left}=\frac{100}{1.5}\text{cm}$

$\therefore$ Magnification of the telescope is:

$m=\frac{f_{\circ }}{f_e}$

For astronomical telescope, focal length of objective lens, $\left(f_{\circ }\right)$ is greater than the eyelens $\left(f_e\right)$.

$\Rightarrow f_{\circ }=f_{left}=\frac{100}{1.5}\text{cm}$

$\text{and}{f}_e=f_{right}=\frac{100}{3}\text{cm}$

$\therefore m=\frac{100}{1.5}\times \frac{3}{100}=2$

Hence, $\left(D\right)$ is correct answer.

Note : For a compound microscope, the focal length of the objective lens is greater than the focal length of the eye piece.