Question

An optical instrument used for angular magnification has a 25 D objective and 20 D eyepiece. The tube length is 25 cm when the eye is least strained. (a) Whether it is a microscope or a telescope? (b) What is the angular magnification produced?

Answer 1

Answer:

more than 30 to 40 zooms not enough

Answer 2

(a) Whether it is a microscope or a telescope : It is a microscopic

(b) The angular magnification produced : m = 20 cm.

Explanation:

Step 1 :

This optical method has

$f_{0}=\frac{1}{25 D}=4 \times 10^{-2} \mathrm{m}=4 \mathrm{cm}$

$f _e=\frac{1}{20 D}=0.05 \mathrm{m}=5 \mathrm{cm}$

The Tube of  length = 25 cm (Standard customisation)

From Step 1 We can say,

(a) Whether it is a microscope or a telescope :

The unit shall be microscopic as $\mathrm{f}_{\mathrm{0}}<\mathrm{f}_{\mathrm{e}}$.

(b) The angular magnification produced :

The image generated by the objective should be on the focal plane of the eye piece since the final image is created at infinity.

So gap to picture for objective = $v_0$ = 25- 5 =20 cm  

We know that lens formula,

$\frac{1}{v_{o}}-\frac{1}{u_{o}}=\frac{1}{f_{o}}$

$\frac{1}{v_{o}}-\frac{1}{f_{o}}=\frac{1}{u_{o}}$

$\frac{1}{u_{o}}=\frac{1}{20}-\frac{1}{4}$

$\frac{1}{u_{o}}=\frac{1-5}{20}$

$\frac{1}{u_{o}}=\frac{-4}{20}$

$\frac{1}{u_{o}}=\frac{-1}{5}$

$u_{o}=-5$

So angular magnification,

$m=-\frac{v_{0}}{u_{o}} \times \frac{D}{f_{e}}$

Taking D = 25 cm

$m=-\frac{20}{-5} \times \frac{25}{5}$

m = 4 × 5  

m = 20 cm .