Question

An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

Answer 1

Focal length of the objective lens, f_o = 1.25 cm

Focal length of the eyepiece, f_e = 5 cm

Least distance of distinct vision, d = 25 cm

Angular magnification of the compound microscope = 30X

Total magnifying power of the compound microscope, m = 30

The angular magnification of the eyepiece is given by the relation:

`"m"_"e" = (1 + "d"/"f"_"e")`

= `(1 + 25/5)`

= 6

The angular magnification of the objective lens (m_o) is related to m_e as:

m_o m_e = m

m_o = `"m"/"m"_"e"`

= `30/6`

= 5

We also have the relation:

`"m"_"o" = ("Image distance for the objective lens" ("v"_"o"))/("Object distance for the objective lens" (-"u"_"o"))`

`5 = "v"_"o"/-"u"_"o"`

∴ v_o = −5u_o ................(1)

Applying the lens formula for the objective lens:

`1/"f"_"o"  = 1/"v"_"o" - 1/"u"_"o"`

`1/1.25  = 1/(-5"u"_"o") - 1/"u"_"o"`

`1/1.25 = (-6)/(5"u"_"o")`

u_o = `(-6)/5 xx 1.25`

∴ u_o = −1.5 cm

And v_o = −5u_o

= −5 × (−1.5)

v_o = 7.5 cm

The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.

Applying the lens formula for the eyepiece:

`1/"v"_"e" - 1/"u"_"e" = 1/"f"_"e"`

Where,

v_e = Image distance for the eyepiece = −d = −25 cm

u_e = Object distance for the eyepiece

`1/"u"_"e" = 1/"v"_"e" - 1/"f"_"e"`

= `(-1)/25 - 1/5`

= `- 6/25`

∴ u_e = −4.17 cm

Separation between the objective lens and the eyepiece = `|"u"_"e"| + |"v"_"o"|`

= 4.17 + 7.5

= 11.67

Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.