(i) the telescope is in normal adjustment, (ii) the final image is formed at the least distance of distinct vision.

Question

(a) Draw a ray diagram of compound microscope for the final image formed at least distance of distinct vision?

(b) An angular magnification of 30X is desired using an objective of focal length 1.25 cm and an eye piece of focal length 5 cm. How will you set up the compound microscope for the final image formed at least distance of distinct vision?

OR

(a) Draw a ray diagram of Astronomical Telescope for the final image formed at infinity.

(b) A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. Find the magnifying power of the telescope for viewing distant objects when

(i) the telescope is in normal adjustment,

(ii) the final image is formed at the least distance of distinct vision.

Answer 1

(a) Diagram of Compound Microscope for the final image formed at D:

(b) m_o = 30, f_o = 1.25 cm, f_e = 5 cm

when image is formed at least distance of distinct vision,

D = 25 cm

Angular magnification of eyepiece

Total Angular magnification, m = m_om_e

⇒ m_o = \(\frac{m}{m_e}=\frac{30}{6}\) = 5

As the objective lens forms the real image,

m_o = \(\frac{\text{v}_o}{u_o}\) = −5

⇒ v_o = −5u_o

using lens equation, u_o = −1.5 cm,

v_o = −5 × (−1.5)cm = +7.5 cm

Given v_e = −D = −25 cm, f_e = +5 cm, u_e = ?

using again lens equation u_e = \(\frac{25}{6}\)

Thus, object is to be placed at 1.5 cm from the objective and separation between the two lenses should be

L = v_o + Iu_el = 11.67 cm

OR

(a) Ray diagram of astronomical telescope when image is formed at infinity.

(b) (i) In normal adjustment :

Magnifying power.

m = f_o/f_e = (140/5) = 28

(ii) When the final image is formed at the least distance of distinct vision (25 cm) :