Question

(i) Draw a ray diagram showing the image formation by a compound microscope. Obtain expression for total magnification when the image is formed at infinity.

(ii) How does the resolving power of a compound microscope get affected, when

(a) Focal length of the objective is decreased.

(b) The wavelength of light is increased? Give reasons to justify your answer.

Answer 1

(i)

Magnification by objective lens = \(\frac{tan\,β}{tan\,α}\)

tan β = \(\frac{h'}{L}=\frac{h}{f_o}\)

\(\frac{h'}{L}=\frac{h}{f_o}\) = (where L = The distance L, i.e., the distance between the second focal point of the eyepiece is called the tube length of the compound microscope.)

Eyepiece will act as simple microscope; hence we may use the formula of magnification by simple microscope for normal adjustment.

m_e = \(\frac{D}{f_e}\)

Total magnification, m = m_o × m_e

= \(\frac{L}{f_o}\times \frac{D}{f_e}\)

(ii) \(d_{min}=\frac{1.22fλ}{D}\)

(a) From the equation, it is clear that resolving power increases when the focal length of the objective is decreased.

This is because the minimum separation, d_min decrease when f is decreased.

(b) Resolving power decreases when the wavelength of light is increased. This is because the minimum separation, d_min increase when λ is increased.