Question

What is the difference in the construction of an astronomical telescope and a compound microscope ? The focal lengths of the objective and eyepiece of a compound microscope are 1.25 cm and 5.0 cm, respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 when the final image is formed at the near point.

Answer 1

A telescope has objective lenses producing long focal lengths, while a microscope has objective lenses producing short focal lengths.Telescopes and microscopes also substantially differ in the diameters of their lenses.

Given, $$f_0 = 1.25 cm, \, f_c = 5 cm $$

angular magnification $$= 30$$

The magnification produced is given by :

(Magnification of eye piece)

$$m_e = \dfrac{d}{f_e} = \dfrac{25}{5} = 5$$

Now,

Total magnification $$= m = m_0 \times m_e$$

$$\therefore m_0 = \dfrac{30}{5} = 6$$

Since the real image is formed by objective lens,

$$m_0 = \dfrac{v_0}{u_0} = -6$$

So, $$v_0 = - 6 u_0$$

Now, $$\dfrac{1}{v_0} - \dfrac{1}{u_0} = \dfrac{1}{f_0}$$

Substitute the values -

$$\dfrac{1}{-6 u_0} - \dfrac{1}{u_0} = \dfrac{1}{1.25}$$

$$\Rightarrow u_0 = \dfrac{-7 \times 1.25}{6}$$

$$= -1.46 cm$$

(object is at $$1.46 cm$$ from the objective lens)