Question

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?

Answer 1

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

`"m"_"e" = (1 + "D"/"f"_"e") = 1 + 25/5 = 6`

Total Angular magnification,

`"m" = "m"_"o" "m"_"e" => "m"_"o" = "m"/"m"_"e" = 30/6 = 5`

As the objective lens forms the real image,

`"m"_"o" = "v"_"o"/"u"_"o" = -5 => "v"_"o" = -5"u"_"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 = `25/6`

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

L = v_o + Iu_e l = 11.67 cm