Question

(a) Draw a labelled ray diagram of a compound microscope. 

(b) Derive an expression for its magnifying power. 

(c) Why is objective of a microscope of short aperture and short focal length? Give reason

Answer 1

(a) Ray diagram of a compound microscope.

(b) When the final image is formed at the least distance of distinct vision,

\(m = -\frac{v_0}{u_0} (1 + \frac D {f_e})\)

For the image formed at infinity,  u_e = f_e

and \(m = -\frac{v_0}{u_0}. \frac D {f_e}\)

By making focal length of the objective small, the magnifying power can be increased.

(c) The microscope is used in sighting small objects situated nearby. If aperture is large then light ray coming from object expands. So intensity becomes small. If aperture is small then light coming from object expends in small aperture, as a result object is seen more bright. So, aperture is kept small.

Answer 2

(a) Labelled diagram of compound microscope.
The objective lens form image A' B' near the first focal point ofeyepiece.

(b) Angular magnification of objective lens m₀ = linear magnification h'/h

 

where L is the distance between second focal point of the objective and first focal point of eyepiece. If the final image A'' B'' is formed at the near point.

Angular magnification m_e = (1+ D/f_e)

If the final image A'' B'' is formed at infinity, then angular magnification me =D/f_e . . . (2)
Thus, total magnification of the compound microscope

M = m₀ x m_e

= L/f₀ x D/f_e

(c) Aperture and focal length increase or decrease the resolving power of the compound microscope.
Resolving power of microscope is given by

R.P. = (2n sinθ)/(1.22λ)

(i) On decreasing the aperture (diameter) of the objective lens, value of sin q decreases, and hence resolving power decreases.
(ii) On decreasing the focal length of the objective lens, value of sin q increases and hence resolving power increases.